
'Mi 



IE TRIGONOMETRY 






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11 



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Class. 
Book. 



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Copyright N°, 



COMRIGHT DEPOSIT. 



ELEMENTS 



OF 



PLANE TRIGONOMETRY 



richard j. Mccarty 

MEMBER, AMERICAN SOCIETY OP CIVIL ENGINEERS 
MEMBER, AMERICAN SOCIETY OF MECHANICAL ENGINEERS 



ILLUSTRATED 



AMERICAN TECHNICAL SOCIETY 
4 CHICAGO 

1920 



«$* 



COPTBIGHT, 1920, BY 

AMERICAN TECHNICAL SOCIETY 



COPYRIGHTED IN GREAT BRITAIN 
ALL RIGHTS RESERVED 



©CU576530 
SEP 14 1920 



PREFACE 

THE purpose of this treatise is to develop the principles 
of Plane Trigonometry in the simplest and most natural 
manner. 

For convenience of reference the geometrical principles 
upon which Plane Trigonometry is based are clearly set forth. 

Ratios of Angular Measure, for the sake of simplicity and 
accuracy, are consistently treated as abstract numbers since 
that is their nature. 

Conventions of Algebraic Signs are treated so as to show 
that their purpose is to make generally true formulas which 
otherwise would be true only in particular cases. 

In order not to distract the mind with unnecessary theory, 
the principles are developed in ascending order with especial 
reference to immediate application. For this reason loga- 
rithms are omitted, since they are not necessary to a knowledge 
of trigonometry and require a tedious digression at the very 
point at which the science may be made interesting by practice. 
To compensate for this omission round numbers are used, the 
data otherwise simplified, and the arithmetical work made to 
compare favorably with that involved in the use of logarithms. 
In this way the problems are made less formidable and their 
solution more convincing and instructive. In addition to this 
the student is encouraged by illustrative examples taken from 
actual practice. 

Having developed the principles of Plane Trigonometry 
with reference to their immediate application, the treatise is 
extended in the final chapters in order to broaden the view and 
prepare the way for the study of Spherical Trigonometry and 
higher branches of mathematics. 



&faK< 






aA ^f~ 



Kansas City, Missouri March 12, 1920. 



CONTENTS 



CHAPTER I 
MEASUREMENT OF DISTANCE AND DIRECTION 

PAGE 

Direct measurement 1 

Instruments for measuring distance 1-3 

Instruments for measuring angles 3-4 

Indirect measurement 5 

The plane triangle 5 

Geometrical properties and relations of plane triangles 5-7 

CHAPTER II 

DEFINITIONS 

Plane trigonometry 8 

Ratios of angular measure 8 

Relations of the ratios to acute plane angles 9 

Names of the ratios 9 

Relative changes in ratios and angles 10 

Limits of the ratios 10 

Relations among ratios for same angle . 11 

Relations of ratios for complementary angles 12 

CHAPTER III 

TABLES OF RATIOS OF ANGULAR MEASURE 

General description of tables 13 

Use of the tables 13-16 

CHAPTER IV 

THE RIGHT PLANE TRIANGLE 

Formulas for solution of right plane triangles 17 

Conditions of solution 17 

Solution of right plane triangles 

Given hypotenuse and one acute angle 18 

Given one right side and adjacent acute angle 19 

Given one right side and opposite angle 20 

Given hypotenuse and one right side 21 

Given the two right sides. 7 22 

Miscellaneous examples 23-24 



CONTENTS 

CHAPTER V 
CONVENTIONS OF ALGEBRAIC SIGNS 

PAGE 

Reason for using conventions of algebraic signs 25 

Dual significance of the signs + and — 25 

Practical illustrations 26-27 

Statement of conventions as applied to oblique plane triangles 27 

CHAPTER VI 

RATIOS OF ANGULAR MEASURE FOR OBTUSE ANGLES 

Application of the conventions to ratios for obtuse angles 28-29 

Application of the tables to ratios for obtuse angles 29 

CHAPTER VII 

THE OBLIQUE PLANE TRIANGLE 

Formulas for solution of oblique plane triangles 

Formulas of the sine 30 

Formulas of the cosine 31 

Formulas of the tangent r. 32 

Formulas of the perimeter 33 

CHAPTER VIII 

SOLUTION OF OBLIQUE PLANE TRIANGLES 

Conditions of solution 34 

Given one side and two adjacent angles 34 

Given one side, one adjacent and one opposite angle 35 

Given two sides and the angle opposite the greater given side 36 

The anomalous case 37 

Given two sides and their included angle r 

Solution by cosine 38 

Solution by tangent 39 

Given three sides 

Solution by cosine 40 

Solution by perimeter 41 

Miscellaneous illustrative examples 42-45 

CHAPTER IX 

MISCELLANEOUS PROPERTIES AND RELATIONS OF 
PLANE TRIANGLES 

Various relations among sides and angles 46-49 



CONTENTS 

CHAPTER X 

CALCULATION OF AREAS 

Area of plane triangle page 

By sine 50 

By perimeter 51 

Area of regular polygon 51 

CHAPTER XI 

RELATIONS OF RATIOS FOR VARIOUS ACUTE ANGLES 

Ratios for sum of two acute angles 52 

Ratios for difference between two acute angles 53 

Ratios for sum and difference of two acute angles 54 

Ratios for twice an angle 54 

Ratios for half an angle 54 

Ratios for whole and half angles 55 

Ratios for multiple angles 56-57 

Approximate relations of ratios for small angles 58 

CHAPTER XII 
CALCULATION OF RATIOS 

Calculation of particular ratios 59-61 

Calculation of tables 61 

CHAPTER XIII 
RATIOS OF ANGULAR MEASURE FOR ANY ANGLE 

Application of conventions of algebraic signs to angles 62 

General statement of conventions of algebraic signs as applied to lines 

and angles 63 

Algebraic signs of ratios for angles greater than 180° 64-65 

Ratios for negative angles 65 

Ratios for angles greater than 360° 65 

Generalization of formulas f 66-67 

CHAPTER XIV 

General theory of circular measure 68-70 

Circular measure and ratios of angular measure 70 

Relations for small angles 70 

Relations for small increases in angles 71-72 

Calculation of the ratios from circular measure 72-74 



CONTENTS 
CHAPTER XV 

PAGE 

Inverse ratios of angular measure, a 75 

Relations for small increases in ratios 76 

Calculation of circular measure from the ratios 77-78 

CHAPTER XVI 

SUPPLEMENT 

Secant and cosecant 79 

Versed sine and coversed sine 80 



TABLES OF NATURAL SINES, COSINES, TANGENTS, 
AND COTANGENTS 



ELEMENTS OF 
PLANE TRIGONOMETRY 



CHAPTER I 

INTRODUCTION 
MEASUREMENT OF DISTANCE AND DIRECTION 

1. The distance between two points is determined by the 
length of the straight line joining them. 

2. The direction of a straight line is determined by the mag- 
nitude of the plane angle which it makes with some other straight 
line of known position in the same plane. 

Direct Measurement 

3. The direct measurement of a distance is the well-known 
process of finding by actual trial, with some suitable instrument, the 
number of times it contains a definite unit of measure. 

4. The instruments for measuring distance are scales, rules, 
rods, tapes, and chains. 



1 j j 1 1 > 1 1 1 i i ii 1 1 1 |i 1 1 1 i i 1 1 i pi i tj i 1 1 n 1 1 1 ij 1 1 1 1 1 > 1 1 i j - nrrfj i ' t^m 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 i j 1 1 1 1) 1 1 m ii i m 1 1 1 1 i,y in|i 1 1 1 1 i i "|»; ' '.u 



UlftlftlftlftlffiimftlftlftlftlffilrtlftlftlftlftlfllffilftlffilffilffilffilffilftliTili-FiliTilfflliT 



r- - ■ . ,, .. u-.a.sr-B - Jin* lv. ' ' ""W^tC* 1 " wwwwt'/iw' _-~-,. J , 



Typical Engineer's Flat and Triangular Scales 
Courtesy of Keuffel & Esser, New York City 




Folding Rule 
Courtesy of Keuffel is Eaaer, New York City 

In land surveying where the unit of area is the acre, what is 
known as Gunter's chain is used for convenience in calculation. 



PLANE TRIGONOMETRY 




This chain is 66 feet long and consists of 
100 links. Its advantage is that 10 square 
chains of this length equal 1 acre.. 

In railroad surveys and in similar 
work, the Engineer's chain is generally 
used. This chain is 100 feet long and 
consists of 100 links. 




Typical Flat Steel Tape 
Courtesy of Keuffel & Esser, New York City 



Philadelphia Leveling Rod 

with Target 

Courtesy of Keuffel & Easer, 

New York City 




Gunter's Chain 
Courtesy of Keuffel & Esser, New York City 



INTRODUCTION 




Engineer's Tape Chain 
Courtesy of Keuffel & Esser, New York City 



5. The direct measurement of a direction or of an angle is gen- 
erally made by means of an instrument having a circle or the arc of 
a circle graduated to degrees and fractions of a degree. 




Protractor 
Courtesy of Keuffel & Esser, New York City 



Surveying Compass 
Courtesy of Keuffel & Esser, New York City 



The different varieties of these instruments are protractors, 
compasses, transits, and sextants. The instrument generally used 
in Surveying and Engineering is a transit measuring angles to the 
nearest minute. 



PLANE TRIGONOMETRY 




Engineer's Transit 
Courtesy of Keuffel & Esser, New York City 



INTRODUCTION 5 

Indirect Measurement 

6. The indirect measurement of a distance or of a direction 
consists in deriving the number of units of measure it contains from 
other distances and directions that have been measured. 

7. The plane triangle is the natural basis for indirect meas- 
urements of distance and direction. ! 

This is because the Geometrical Properties and Relations of 
Plane Triangles are such that: 

8. Any required distance may be made a side and any 
required direction may be made an angle of some plane triangle. 

9. Any unknown angle of a plane triangle may be determined 
when the other two angles are known. 

10. Any unknown side or angle of any plane triangle may be 
determined when certain other sides and angles of the triangle are 
known. 

GEOMETRICAL PROPERTIES AND RELATIONS OF PLANE TRIANGLES 

11. The Geometrical Properties and Relations of Plane Tri- 
angles upon which indirect measurements are based may, for con- 
venience and reference, be stated as follows : 

12. The sum of the three angles of any plane triangle is two right 
angles or 180°. ^b 

Let ABC be any plane triangle. 
Then 

A+B+C = 180° 
and A = 180°-(B+C) 

B = 180°-{A+C) 
C = 180°-(A+B) *' 

13. In any plane triangle the greater angle is opposite the greater 
side and the greater side is opposite the greater angle. 

In the plane triangle ABC, let BC 
be greater than AB. 

Then angle A > angle C 

And conversely. Similarly for 
other sides and angles. 





PLANE TRIGONOMETRY 



14. Plane triangles which have two angles of one equal, respec- 
tively, to two angles of the other have all corresponding angles equal 
and all corresponding sides equal or proportional. 



Let ABC and DEF be two 
plane triangles in which A = D 
and B = E. 

Then C=F 



and 
AB : DE = BC : EF=AC : DF 





15. Plane triangles which have two sides of one equal or pro- 
portional to two sides of the other and the angles opposite the greater 
of these sides equal have all corresponding angles equal and all corre- 
sponding sides equal or proportional. 

Let ABC and DEF be two 
plane triangles in which 

AC:DF=CB :FE 

A = D 
and CB>AC FE>DF 
Then B = E C = F 
and AC :DF=AB : BE 

16. Plane triangles which have two sides of one equal or pro- 
portional to two sides of the other and their included angles equal have 
all corresponding angles equal and all corresponding sides equal or 
proportional. 

Let ABC and DEF be two 
plane triangles in which 

AB :DE=AC : DF 

and A = D 

Then B = E C=F 

and AB :DE = CB :FE 




INTRODUCTION 



17. Plane triangles which have all their corresponding sides pro- 
portional have all their corresponding angles equal. 

Let ABC and DEF be two 
plane triangles in which 
AB :DE = BC : EF=AC :DF 

Then 
A=D B=E C=F 

18. The square of the hypotenuse of a right plane triangle is equal 
to the sum of the squares of the other two sides. 

Let AHB be a right triangle. 

Then ZB 2 =M 2 +Zff 2 




and 



BH 2 = AB 2 -AH 2 
AH 2 =AB 2 -BH 2 




19. In any plane triangle, the square of any required side is 
equal to the sum of the squares of the other two sides minus twice the 
product of one of those sides by the projection of the other upon that 
side if the angle opposite the required side is acute, but plus that product 
if the angle is obtuse. 

Let ABC be a plane triangle and BC be the required side. 
Draw BH perpendicular to CA or CA prolonged. 





*l 



When A is acute 



BC 2 =A(?+AB 2 -2(ACXAH) 



When A is obtuse 
W=AC 2 +A1?+2(ACXAH) 
20. The propositions just given may be applied to Indirect 
Measurement of Distance and Direction either by maps and draw- 
ings according to Geometry or by calculation according to the prin- 
ciples of Trigonometry. 



CHAPTER II 

DEFINITIONS 

21. Plane Trigonometry treats of the calculation of the 
unknown sides and angles of plane triangles from sides and angles 
that are known. 

22. The three straight lines and the three angles that form a 
plane triangle are its parts. 

23. The calculation of all the unknown from the known parts 
is the solution of the triangle. 

24. In order to establish between the sides and angles of plane 
triangles the arithmetical and algebraical relations required for calcu- 
lation it is necessary to use certain abstract quantities known as 
ratios of angular measure. 

RATIOS OF ANGULAR MEASURE FOR ACUTE ANGLES 

25. The ratios of angular measure for acute angles are quo- 
tients obtained from the sides of the right plane triangle formed by 
drawing from a point on one side of the angle a perpendicular to 
the other side. Thus: 



Let VAZ be an acute plane angle. 
From any point B on AV draw BH 
perpendicular to AZ. 

Let a, b, and h denote the number 
of times BH, AH, and AB, respec- 
tively, contain a common unit of 
measure. 

Then the ratios of angular measure of the acute angle 
VAZ are 

a_ b a b h h _ a , 1 6_ 

A o a o a h h 




RATIOS FOR ACUTE ANGLES 




RELATIONS OF RATIOS TO ACUTE PLANE ANGLES 

26. Let VAZ and YDX be two 
acute plane angles. From any point 
B on AV draw BH perpendicular to 
AZ and from any point F on DY 
draw FE perpendicular to DX. 

Let VAZ=YDX 

Then, by 14, all corresponding 
sides of the right triangles A HB and 
DEF are proportional. 

Whence, equal acute angles have 
equal corresponding ratios. 

27. Let the ratio of the hypotenuse to either right side of the 
triangle AHB equal the corresponding ratio in the triangle DEF. 

Then, by 15 angle VAZ = angle YDX 
Let the ratio of the right sides in one triangle equal the cor- 
responding ratio in the other. 

Then, by 16 angle VAZ = angle YDX 

Whence, equal corresponding ratios have equal acute angles. 

NAMES OF RATIOS OF ANGULAR MEASURE 

28. To distinguish the ratios from one another, each has been 
named according to the position of its lines with respect to the 
angle to which it relates. Thus, in the right triangle AHB having 
sides and angles as shown, 



sine of 



cosine of 



A side opposite . a 

A = - — or sin A = — 

hypotenuse h 

A side adjacent . 

A = - or cos A = „ 

hypotenuse h 

- A side opposite A a 

tangent of A = —. — yf or tan A=— 

side adjacent b 

« A side adjacent b 

cotangent ot A = —. — : — or cot A = — 

side opposite a 

29. The other ratios indicated 
in 25 are not necessary for the & £~ 

solution of plane triangles. A general discussion of these ratios is 
given in the supplement. 




10 



PLANE TRIGONOMETRY 



RELATIVE CHANGES IN RATIOS AND ANGLES 

30. Let the acute angle 
VAZ = A and let it be in- 
creased by a small angle 
VAQ = x, so that A+x will 
be acute. 

With A as a center, de- 
scribe the arc of a circle CBF. 
Draw BH and FE perpendic- 
ular to AZ. 




Then 

sin (A+x)- 



FE>BH and AE<AH and 

AE 



FE^ . A BH t a \ \ 

"AF >SmA = AB C0 ^ A+X ^ 



A AH 

AF <C ° SA = AB 



, A . N FE , BH 

tan (A+x) =—^>tan A=—- 

AEi An. 



cot(^+x)=^|<cot^=|| 



Therefore, the sine and tangent of an angle increase and the 
cosine and cotangent decrease as the angle is increased. 

Similarly it may be shown that the sine and tangent diminish and 
that the cosine and cotangent increase as an acute angle is diminished. 



LIMITS OF RATIOS 



31. Let YAX be a right angle, 
the arc of a circle DFC; let BH be 
the perpendicular from a moving point 
B on the arc to AX and let BAC = A. 

When AB coincides with AX, 
A = Q, BH = 0, and AH = AB. 

When AB coincides with AY, 
A =90°, BH = AB, and AH = 0. 

Whence 

BH 



With iasa center, describe 

Y 



sm 



cos 0° 



°° = AB=AB = Zer0 

AH AB 
: _ = _ = umty 

+ no BH 

tan ° = AH = AB = Zer ° 

cot 0° = — = — = infinity 



sin 




L-+- 



H 



nno BH AB . 
90 = AB = AB = Umty 



AH 



cos 90° = -TT7 = -r^ = zero 



AB 

tan 90° = ^ = 
AH 

cot 90° = 4S = 





AB 

AB 







= infinity 



BH AB 



= zero 




RATIOS FOR ACUTE ANGLES 11 

RELATIONS AMONG RATIOS FOR SAME ANGLE 

32. In the right triangle AHB, we have from 18 

a 2 +b 2 = h 2 £^f\* 

Therefore — - + — = 1 
h 2 h 2 

But by definitions given in 28 

— = sin^4 and — = cos^4 
h h 

Whence 

sin 2 A+cos 2 A= 1 jf 

or, the square of the sine plus the square of the cosine of an angle is 
equal to unity. 

Whence also sin 2 A = 1 — cos 2 A 

and cos 2 A = 1 — sin 2 J. 

33. Bv definitions in 28, tan A = — 

b 

But by identity -r = -rX-r =—-i—r 
b h b a h 

Whence tan A = - 

cos A 

or, the tangent of an angle is equal to its sine divided by its cosine. 

34. By definitions in 28, cot A= — 

a 

But by identity — = t~ X — == ~r~ i '~r 
a n a h h 

Whence cot A = 

sin A 

or, the cotangent of an angle is equal to its cosine divided by its sine. 

35. By definition 28, tan A = — and cot A = — 

b a 

7 

Multiplying tan A cot ^4 = — X — = 1 

b a 

Whence tan A = and cot A = - 



cot A tan A 

or, the tangent and cotangents are reciprocals of each other. 



12 



PLANE TRIGONOMETRY 



EXAMPLES FOR PRACTICE 

36. Verify the following equations: 

1. sm 2 A = (l — cos^4)(l+cos ^4) 

2. cos 2 ^ = (l-sin^l)(l+sin^) 

3 



4. tan ^4+cot A = - 



1 



sin 2 J. 9A 1 — cos 2 ^4 

- = tan 2 ^4 



sin A cos A 

5. sin 4 J. — cosM = sin 2 J. — eos 2 ^ 

6. tan A cos A sin A+qo&A = 1 



1 — sin 2 J. cosM. 

7. (sin 2 ^-cos 2 ^) 2 = l -4sin 2 ^cos 2 ^ 



RELATIONS OF RATIOS FOR COMPLEMENTARY ANGLES 

37. Two angles are complementary when their sum is 90°. 

38. In the right triangle AHB 
with sides and angles as shown, we 
have from 12 

A+B+H = 180° 

Hence ^4+5 = 90° 

A = 90°-B 

and B = 90°-A 

Also by definition 28 



sin ^4=— = cos B 

h 




cos .^4=— = sin B 
h 



tan A = 



a 



-cotB 



cot A= — = tan B 
a 



Whence 



sin A = cos (90° -A) 
tan A = cot (90° -A) 



cos A = sin (90° -A) 
cot A = tan (90° -A) 



EXAMPLES FOR PRACTICE 

39. 1. Verify these relations : 

Sin 63° 24' = cos 26° 36' cos 36° 34' = sin 53° 26' 
Tan 59° 16' = cot 30° 44' cot 19° 12' = tan 70° 48' 

2. Given sin A = cos 2 A. Find A Ans. 

3. Given sin 2A = cos 3A. Find A Ans. 

4. Given tan A = cot (18°+-4). Find A 

5. Given cot A =tan (62° +A). Find A 



,4 = 30° 

^4 = 18° 

Ans. .4=36° 

Ans. .4 = 14° 



CHAPTER III 

TABLES OF RATIOS OF ANGULAR MEASURE 

40. The mathematicians, by means of relations that exist 
among the ratios for the same and for different angles, have cal- 
culated and compiled various tables from which the numerical 
values of ratios for a given angle and the angle corresponding 
to the numerical value of a given ratio may be found. These 
tables give the numerical values of the ratios at different angular 
intervals and to different numbers of decimal places according to 
the degree of accuracy required. The numerical ratios given in 
such tables are known as Natural Ratios to distinguish them 
from other quantities that have been derived from them. 

41. The annexed tables give the numerical values of natural 
sines, cosines, tangents, and cotangents at intervals of 10' for 
angles from 0° to 90° and to four and five places of numerals. 

USE OF TABLES 

42. To find either ratio for any given angle shown in this 
table. First find the given degrees and then, in the next column 
nearest below, find the given minutes. On the line of given 
minutes and in the proper column the required ratio will be 

found. 

Thus: sin 16° 30'= .2840 sin 67° 20'= .9228 

cos 21° 50'= .9283 cos 86° 10'= .0669 

tan 24° 30'= .4557 tan 73° 50' = 3.4495 

cot 28° 20' = 1.8546 cot 78° 40'= .2004 

43. To find the angle for a given tabular value of either 
ratio. Find the given value in the proper column. On the same 
line to the left will be the minutes and in the next column nearest 
above will be the degrees. Thus: 

The sine .1965 = sin 11° 20' .9993 = sin 87° 50' 

cosine .9769 = cos 12° 20' .6041 = cos 52° 50' 

tangent .1198 = tan 6° 50' 2.5386 = tan 68° 30' 

cotangent 2.8770 = cot 19° 10' .1405 = cot 82° 00' 



14 PLANE TRIGONOMETRY 

INTERPOLATION 

44. To compile tables showing the ratios for every required 
angle is obviously impracticable. Interpolation is the process 
by which intermediate angles and ratios are calculated from the 
ratios and angles given in a table. This process is based on the 
assumption that, within the tabular intervals, changes in angles 
are proportional to changes in corresponding ratios. Thus: 

Let a=the angular interval of the table 

c = a change of less than a in a given angle 

d = difference between adjacent tabular values of a ratio 

corresponding to the given angle 
r = change in ratio corresponding to change c in angle 

Then a: c:: d: r 

The results under this theory are practically correct for all 
ratios except for tangents when the acute angle is large and for 
cotangents when the angle is small. The angles to which these 
exceptions apply depend upon the tabular intervals and the 
number of decimal places in the table. With tables of 10' inter- 
vals the use of tangents of acute angles greater than 70° and 
cotangents of angles less than 20° should be avoided. 

45. In making interpolations it must be borne in mind that, 
as shown in 30, sines and tangents increase and cosines and cotan- 
gents decrease as the angle increases. 

PROBLEMS FOR ILLUSTRATION 

46. 1. To find sin 29° 13'. -- ' 

For next less angle, sin 29° 10' = .4874 
and for next greater, sin 29° 20' = .4899. Here, increase in 

ratio for increase in angle of 10' is .0025, or for 1' is .00025. 

Given angle -29° 10' =29° 13' -29° 10' = 3'. Increase in ratio for 
increase of 3' in angle = .00025X3 = .0007, nearly. 

Whence sin 29° 13' = sin 29° 10' +.0007 = .4881 

Examples. Verify sin 24° 46' = .4189 
sin 77° 19' = .9756 



TABLES OF RATIOS OF ANGULAR MEASURE 15 

2. To find cos 41° 24' 

For next less angle, cos 41° 20' = .7509 
and for the next greater, cos 41° 30' = .7490. Here, decrease in 
ratio for increase in angle of 10' is .0019, or for 1' is .00019. 

Given angle-41°20'=41°24'-41°20'=4'. Decrease in ratio for 
increase of 4' in angle = .00019X4 = .0008, nearly. 

Whence cos 41° 24' = cos 41° 20' - .0008 = .7501 

Examples. Verify cos 19° 15' = .9441 
cos 66° 18' = .4019 

3. To find tan 27° 16' 

For next less angle, tan 27° 10' = .5132 
and for next greater, tan 27° 20' = .5169. Here, increase in 

ratio for increase in angle of 10' is .0037, or for 1' is .00037. 

Given angle-27° 10'=27° 16'-27° 10' = 6'. Increase in ratio for 
increase of 6' in angle = .00037X6 = .0022, nearly. 

Whence tan 27°16'=tan 27° 10'+.0022 = .5154 

Examples. Test, tan 44° 57' = .9983 
tan 58° 12' = 1.6128 

4. To find cot 72° 43' 

For next less angle, cot 72° 40' = .3121 
and for next greater, cot 72° 50' = .3089. Here, decrease in 

ratio for increase in angle of 10' is .0032, or for 1' is .00032. 

Given angle -72° 40' = 72° 43' -72° 40' =3'. Decrease in ratio for 
increase of 3' in angle = .00032X3 = .0010, nearly. 

Whence cot 72°43' = cot 72°40'-.0010 = .3111 

Examples. Test, cot 27° 14' = 1 .9430 
cot 69° 47'= .3683 

5. To find the angle for the sine .6972 
For next less ratio, .6968 = sin 44° 10' 

and for next greater, .6988 = sin 44° 20'. Increase in angle 

for increase in ratio of .0020 is here 10', or for .00020 is 1'. 

Given sine - sin 44° 10' = .6972 - .6968 = .0004'. Increase in angle for 

increase of .0004 in ratio = '— — — Xl'=2'. 

.UUU.Z 

Whence angle for sine .6972 = 44° 10'+2' = 44° 12' 
Examples. Test, angle for sine .2742 = 15° 55' 
angle for sine .8502 = 58° 14' 



16 PLANE TRIGONOMETRY 

6. To find the angle for the cosine .3101 
For next less ratio, .3090 = cos 72° 00' 

and for next greater, .31 18 = cos 71° 50'. Decrease in angle 

for increase in ratio of .0028 is here 10', or for .00028 is 1'. 
Given cosine-cos 72° 00' = .3101 -.3090 = .0011. Decrease in angle 

for increase of .0011 in ratio = * noQ X 1/ = 4', nearly. 

.00028 

Whence angle for cosine .3101 = 72° 00'-4' = 71° 56' 

Examples. Test, angle for cosine .8425 = 32° 36' 

angle for cosine .5515 = 56° 32' 

7. To find the angle for the tangent .9451 
For next less ratio, .9435 = tan 43° 20' 

and for next greater, .9490 = tan 43° 30'. Increase in angle 

for increase in ratio of .0055 is here 10', or for .00055 is 1'. 
Given tangent-tan 43° 20' = .9451 -.9435 = .0016. Increase in 

angle for increase of .0016 in ratio = : — — — Xl' = 3', nearly. 

.000o5 

Whence angle for tangent .9451 =43° 20'+3' =43° 23' 

Examples. Test, angle for tangent .8371 =39° 56' 

angle for tangent 1.7832 = 60° 43' 

8. To find the angle for the cotangent .4564 
For next less ratio, .4557 = cot 65° 30' 

and for next greater, .4592 = cot 65° 20'. Decrease in angle 

for increase in ratio of .0035 is here 10', or for .00035 is 1'. 
Given cotangent -cot 65° 30' = .4564 -.4557 = .0007. Decrease in 

angle for increase of .0007 in ratio = ' „ r X 1' = 2'. 

.00035 

Whence angle for cotangent .4564 = 65° 30' -2' = 65° 28' 

Examples. Test, angle for cotangent .1874 = 79° 23' 

angle for cotangent 1.5253 = 33° 15'. 

47. For ordinary problems in Surveying and Engineering the 
Tables given will serve but where extreme accuracy is required, and 
particularly in Geodetic Surveys, Navigation and Astronomy, tables 
giving values to a greater number of decimal places and to smaller 
intervals are necessary. 



CHAPTER IV 



THE RIGHT PLANE TRIANGLE 
FORMULAS FOR SOLUTION OF RIGHT PLANE TRIANGLES 

48. General formulas for the solution of right plane triangles 
are established as follows: 

Let A HB be any right plane triangle with angles and sides as 
shown. 
Then by Geometry, 12 

A+B = 90° 
A = 90°-B 
and B = 90°-A 

By definition 28 

sm A=-r = cosB, tan A = -r- = cot B 

h h 

cos ^4=-T- = sin5, cot A = — = tanB 
h a 




Whence 



a = h sin A = h cos B = b tan A = b cot B 
b = h cos A = h sin B = a cot A = a tan B 
a a b b 



h = 



sin ^4 cos B sin £ cos A 



SOLUTION OF RIGHT PLANE TRIANGLES 

49, By means of these formulas and the annexed tables, all 
the unknown angles »and sides of any right plane triangle may be 
calculated when, besides the right angle, there are known: 

I. The hypotenuse and an acute angle 

II. One right side and the adjacent acute angle 

III. One right side and the opposite angle 

IV. The hypotenuse and one right side 
V. The two right sides 



18 



PLANE TRIGONOMETRY 



50. Problem I. Given the hypotenuse and one acute angle, 
to calculate the remaining parts. 
From formulas given in 48 

^+£ = 90 

a = h sin A 
a = h cos B 
b = h cos A 
b = h sin B A- 




Examples. 

B, a, and b. 



1. Given h = 1000 ft. and A = 31° 20', to calculate 



B = 90 o -31 o 20' = 58°40' 

sin A = sin 31° 20' = .5200 

cos A = cos 31° 20' = .8542 
a = h sin A = 1000 ft. X .5200 = 520 ft. 
b = h cos A = 1000 ft. X .8542 = 854.2 ft. 

2. Given h = 1000 ft. and B 72° 30', to calculate A, a, and b. 

Ans. ^ = 17° 30'; a = 300.70 ft.; 6 = 953.70 ft. 

3. To prolong the line DA. 




Take some convenient point B from which a line B H can be run 
perpendicular to the direction of DA. Measure the angle A and 
the distance AB. Lay off the angle 5 = 90°— A and the distance 
BH — AB sin A. Then H will be on the line DA and its distance 
from A will be AE — AB cos A. 

(1) Given ^15 = 1000 ft. and ^=26° 10', to calculate B, BE, 
and AH. Ans. B = 63° 50' ; BE = 441 f t. ; AH = 897.5 ft. 

(2) Given ,45 = 1000 ft. and ^ = 39° 50', to calculate B, BE, 
and AE. Ans. B = 50° 10'; B E = 640.6 ft.; A E = 767.9 ft. 



FORMULAS FOR RIGHT PLANE TRIANGLES 19 



51. Problem II. Given one right side and the adjacent acute 
angle of a right plane triangle, to calculate the remaining parts. 

From formulas given in 48 
4+5 = 90° 
a=b tan A 
6 = a tan B 
a 



h = 



h = 




Examples. 

a, and h. 



cos B 

b 
cos A 

1 . Given 6 = 797 ft. and A = 28° 49', to calculate B, 

5 = 90°-28°49' = 61°ll' 
tan A= tan 28° 49' = .5501 
cos A = cos 28° 49' = .8762 

a = 797 ft. X. 5501 =438.4 ft. 

797 ft. 



h = - 



.8762 



= 909.6 ft. 



2. Given 6 = 1000 ft. and ,4 = 28° 50', calculate B, a, and h. 

Ans. 5 = 61° 10* o = 550.5 ft.; h = 1141.5 ft. 

3. Given a = 1000 ft. and B = 48° 20', to calculate A, b, and h. 

Ans. ,4=41° 40'; 6 = 1123.7 ft.; h = 1504.2 ft. 

4. To calculate the distance HB across a river. 

Lay off a conven- 
ient base line HA at 
right angles to HB. 
Measure HA and the 
angle HAB. 

1 . Suppose HA = 
100 ft. and HAB = 78° 
20', calculate HB. 

Ans. H B = 484.3 ft. — H 

2. Suppose HA = 200 ft. and HAB = 73° 50', calculate HB. 

Ans. #5 = 689.9 ft. 

3. Suppose HA = 300 ft. and HAB = 67° 40', calculate HB. 

Ans. HB = 730.3 ft. 




20 



PLANE TRIGONOMETRY 



52. Problem HI. Given one right side and the opposite angle 
of a right plane triangle, to calculate the remaining parts. 

From formulas given in 48 

A + B = 90° 
a = b cot B 
b = a cot A 

sm A 

sm B 

Examples. 1. Given 6 = 123 
ft. and 5 = 48° 42', to calculate A, 
a, and h. 

■4 = 90 o -48°42' = 41°18' 
cot B = cot 48° 42' = .8785 
sin£ = sin48°42' = .7513 

a = 123ft.X.8785 = 108.1ft. 
123 ft. 




h=- 



.7513 



:163.7 ft. 



2. Given 6 = 1000 ft. and 5 = 48° 40', to calculate A, a, and h. 

Ans. ^ = 41° 20'; a = 879.6 ft.; A = 1331.7 ft. 

3. Given a = 300 ft. and ^4 = 19° 30', to calculate B, b, and h. 

Ans. B = 70° 30' ; b = 847.2 ft. ; h = 898.7 ft. 

4. To calculate the distance to the moon. 

Let E be the center of the earth, P a point on its surface, and let 
M be the center of the moon. 




"© 



Astronomers have found that the distance EP = 3960 miles and 
that when the angle EPM = 90°, the angle EM P = 57'. Calculate 
the distance EM. Ans. EM = 238,500 miles, nearly 



FORMULAS FOR RIGHT PLANE TRIANGLES 21 



53. Problem IV. Given the hypotenuse and one right side 
of a right plane triangle, to calculate the remaining parts. 
From formulas given in 48 
,4 + 5 = 90° 

sin A=-j- = cos B 
h 

• » b A 

sin B = ~r = cos A 
n 

hsin A=a = h cos B 
h sin B = b = h cos A 




Examples. 
A, B, and a. 



1. Given h = 1250 ft. and 6 = 767.6 ft., to calculate 



sin £ = ?§^ = .6141=sin37 



53' 



Whence 



1250 
B = 37° 53' and .4 = 52° 7' 
sin^ = sin52°7' = .7893 

a=1250 ft. X. 7893 = 986.6 ft. 

2. Given A = 100 m. and a = 78.98 m., to calculate A, B, and b. 

Ans. 4 = 52° 10'; 5 = 37° 50'; 6 = 61.34 m. 

3. Given h = 1000 ft. and b = 664.8 ft., to calculate A, B, and a. 

Ans. 4=48° 20'; 5=41° 40'; a = 747.00 ft. 

4. Given h = 2000 ft. and a = 1000 ft., to calculate A, B, and b. 

Ans. ^ = 30° 00'; £ = 60° 00'; 6 = 1732.0 ft. 

5. Given h = 1200 m. and 6 = 1165.2 m., to calculate A, B, 
and a. Ans. 4 = 13° 50'; 5 = 76° 10'; a = 286.9 m. 

6. To calculate the area of a rectangle when only one side and 
the diagonal can be found by direct measurement. 




Given 4 #=1841 ft. and .45 = 2000 ft., to calculate the area 
ACBH. Ans. 1,438,557 sq. ft. = 33.02 acres 



22 



PLANE TRIGONOMETRY 



54. Problem V. Given the two right sides of a right plane 
triangle, to calculate the remaining parts. 

From formulas given in 48 B 



A (l n 

tan ^4 = -7- = cot B 
b 

* h 
cot A = — = tan B 
a 



a 



sin A 

a 
cos B 



= A=- 



6 



= h = 



sin B 

b 
cos A 




a b 

Examples. 1. Given a = 632.79 m. and 6 = 1250 m., to calcu- 
late A, B, and h. 

632.79 



tan A = - 



-=.5062=tan26°51' 



Whence 



1250 

^=26° 51' and 5 = 63° 09' 

sin A = .4517 

, 632.79 m. 1/mnn 
h = — -—=, — = 1400.9 m. 



.4517 

2. Given a = 1000 ft. and 6 = 505.9 ft., to calculate A, B, and h. 

Ans. ^ = 63° 10'; 5 = 26° 50'; A=1120.7 ft. 

3. Given a = 500 m. and 6 = 684 m., to calculate A, B, and h. 

Ans. ^ = 36 o 10';B = 53°50 , ;A=847.3m. 

4. Given a = 2877 ft. and 6 = 1000 ft., to calculate A, B, and h. 

Ans. ^4 = 70° 50'; 5 = 19° 10'; A = 3045.8 ft. 

5. To calculate the length AB of a diagonal "eye" bar for a 
bridge span. 




Measured between the centers of the pins, the depth BH of a 
truss = 27.36 ft. and the length HA of a panel =20 ft. 

Ans. AB = 33.89 ft. 



FORMULAS FOR RIGHT PLANE TRIANGLES 23 

MISCELLANEOUS EXAMPLES 

1. A surveyor placed his transit in line with one side of a 
building 100 feet square and found the angle subtended by the 




Center of Instrument^ 



other side to be 5° 20'. Calculate the distance from the center of 
the instrument to the nearest point of the building. 

(Use tangent, see 44.) Ans. 1070.7 ft. 

2. In a plane triangle AHB, H = 90°, a = l ft., 6 = 1 yd.; cal- 
culate A, B, and h. Ans. ^ = 18° 26'; 5 = 71° 34'; A = 3.16 ft. 

3. The angle of ascent of a street grade is 2° 00'. Calcu- 

too'__ 



late the rise for each 100 feet measured along the surface. 

Ans. 3.49 ft. 

4. A flag staff is known to be 100 feet high. A surveyor from 

a point which is on a level with the base observes the angle of 

elevation of the top to be 2° 40'. Calculate the distance from 




the center of the instrument to the base of the flag staff. 

(Use tangent, see 44.) Ans. 2146 ft. 

5. In calculating the grade of a street it is found that the 
ascent is 6.7 feet for each 100 feet measured horizontally. Calculate 
the angle of ascent. Ans. 3° 50' 

6. Estimated from the length of the string, a kite is 1000 feet 
from the observer and its angle of elevation is 43° 40'. Calculate 
the altitude of the kite. Ans. 690.5 ft. 



24 



PLANE TRIGONOMETRY 



7. In a horizontal engine the stroke is 6 feet or crank arm 
3 feet and the connecting rod is 14 feet long. Calculate the angle 




of the connecting rod with the axis of the cylinder when the crank 

arm is vertical. Ans. 12° 22.4' 

8. In running a line, a surveyor, in order to avoid a thick wood, 

turned 45° to the right at a point A and measured 1200 feet to a 










point 0. He then turned 90° to the left and measured 1200 feet to 
a point B on his original line prolonged. Calculate the distance AB. 

Ans. 1697.07 ft. 

9. Viewed from a level prairie the top of a mountain has an 

elevation of 5° 30'. From a point 1000 feet nearer, its elevation is 




5° 50'. Calculate the height of the mountain. 



Ans. 1668 ft. 



CHAPTER V 

CONVENTIONS OF ALGEBRAIC SIGNS 

55. The principles of Plane Trigonometry which so far have 
been developed by means of elementary Algebra and Geometry 
have been shown to be sufficient for the establishment of formu- 
las for the solution of all right plane triangles. 

56. With these principles as the sole basis certain formulas 
for the solution of oblique plane triangles might be established 
by means of the right triangles formed by drawing from the 
vertices of an oblique plane triangle perpendiculars to the 
opposite sides. But on account of differences in algebraic signs, 
such, for instance, as in the two formulas shown in 19, there 
would result in certain cases one set of formulas for acute trian- 
gles and another set for triangles that are obtuse. This would 
leave much to be desired in practice because of the complica- 
tions involved in the use of two sets of formulas for the same 
triangle. 

Therefore, it is important in the interest of simplicity and 
accuracy that these differences in algebraic signs should be 
eliminated. 

57. The elimination of all the algebraic differences indicated 
and the establishment of a single set of formulas which are true 
both for acute and for obtuse plane triangles may be readily 
accomplished by means of what are known as Conventions of 
Algebraic Signs. 

58. Conventions of algebraic signs are based upon the follow- 
ing algebraic principles: 

(1) The signs + and — may be used not only to denote addition 
and subtraction but also and at the same time to express opposite 
directions with respect to each other. 

(2) The signs + and — when used to designate opposite direc- 
tions will make generally true formulas which otherwise would be true 
only in particular cases. 

59. These elegant and important principles admit of the 
following simple illustrations. 



26 PLANE TRIGONOMETRY 

PROBLEMS FOR ILLUSTRATION 

60. Let A and C be towns on the line X'X and suppose a 
man has started from A and gone a given distance on that line. 
Required his distance from C. 

Let a = required distance; b = distance from A to C; and c = dis- 
tance traveled. Also let all distances be regarded as positive and 
let the signs + and — denote only addition and subtraction. 

Suppose the man had gone in the direction of C. 



X 1_* y 



Then a = b-c (1) 

But suppose he had gone in the opposite direction. 

* 1 £ 2 . 2 1 X 

Then a = b+c (2) 

Thus to express his distance from C would require a special 
formula for each particular case. But by the principles of Algebra 

(1) may be written a = 6 — (+c) 

(2) may be written a = b — (—c) 

Now let c be regarded as positive w T hen in the direction of C and 
as negative when in the opposite direction. Then in either case the 
distance of the man from C will be given by the single formula 

a = b—c 

And similarly for any other opposite directions. 

61 • Suppose a man starts from A and goes a given distance 
along a straight course making an angle A with the line X'X. 
Required his distance from a point C on that line. 

Let a = required distance; b = distance from A to C; c = distance 
traveled; and d = projection of c on X'X. Also let all distances be 
regarded as positive and let the signs + and — denote only addition 
and subtraction. 



CONVENTIONS OF ALGEBRAIC SIGNS 



27 



Suppose the man travels along AB making an acute angle A 
with AX. Draw BH perpen- 
dicular to AX. 

Then from 19 



a 2 = b 2 +c 2 -2bXd 



(1) 




X' 




H 



ft 



But suppose he goes along AB f at an obtuse angle A with AX, 
or at an acute angle 180° — A with AX'. 
Then from 19 

a 2 = b 2 +c 2 +2bxd (2) 

Also, by Algebra, from 

(1) a 2 = b 2 +c 2 -2bX(+d) 
and from 

(2) a 2 = b 2 +c 2 -2bX(-d) 

Now let d be regarded as positive when in the direction of C 
and as negative when in the opposite direction. 

Then in either case will the distance a be given by the single 
formula 

a 2 = b 2 +c 2 -2bXd 

62. By virtue of the principles just illustrated and in order 
that the formulas for distance and direction may be generally true, 
mathematicians have adopted the following conventions. 

63. Distances considered 
without regard to distances in the 
opposite direction must be regarded 
as positive. 

64. Distances in opposite 
directions and considered with 
respect to each other must be given 
contrary algebraic signs. __*_= ^ ^ 

65. These conventions are 

applied to the solution of oblique triangles by means of ratios of 
angular measure for obtuse angles. 




CHAPTER VI 



RATIOS OF ANGULAR MEASURE FOR OBTUSE ANGLES 

66. Ratios of angular measure for an obtuse angle are derived 
from the right triangle formed by drawing from any point on 
one side of the angle a perpendicular to the other side prolonged. 

Let the obtuse angle 
XA V = A. From any point B' 
on A V draw B'H' perpendic- 
ular to XA prolonged. 

Then 

AH' 



sin A- 



tan A = 



H'B' 
AB' 

H'B' 
AH' 



cos A = 



AB' 



cotA= AK> 




From AX lay off acute angle XAZ = (180° -A). Also lay off 
AB = AB' and draw BH perpendicular to AX. Then the corre- 
sponding sides of right triangles AHB and A H'B' will be equal. 

By convention of algebraic signs, 63, the distances AB, AB' , 
HB, and HB', having no relation to distances in opposite direction, 
are positive. But by 64, distances AH and A H ', having opposite 
directions, must be taken with contrary signs. 

AB = AB', HB = H'B', and AH'=-AH 



Hence 
Therefore 



sin A- 


H'B' 
' AB'' 


HB 
= AB~" 


cos A 


AH' 
~ AB'' 


-AH 
AB 


tan A- 


H'B' 
~ AH'' 


HB 
-AH 


rnt A - 


AH' 


-AH 



=VE =sin ( 180 °-^) 



-cos (180° -A) 



-= -tan (180° 



H'B' HB 
67. Whence 

sin A= sin (180°-A) 

tanA=-tan(180°-A) 

Also sin (180° -A) = sin A 

tan (180° - A) = -tan A 



= -cot (180° 



-A) 

■A) 



cos A= -cos (180°- A) 
cotA=-cot(180°-A) 
cos(180°-A) = -cosA 
cot (180°-A) = -cot A 



RATIOS FOR OBTUSE ANGLES 29 

68. In all the equations just established, if A is made acute, 
180° — A becomes obtuse and the equations hold true. In either 
case the ratios for the acute angle must be regarded as naturally positive 
as in formulas for right triangles. 

69. The equations given in 67 make it possible to establish 
formulas that are true for both acute and obtuse triangles. They 
also furnish means for determining any ratio for any obtuse angle 
from Tables of Ratios for Acute Angles. 

APPLICATION OF TABLES TO OBTUSE ANGLES 

70. To find the sine of a given obtuse angle A. 
Since from 67 sin A= sin (180° — A) 

Subtract the given angle from 180° and find in the table the sine 
of the resulting acute angle. 

Example. sin 167° 30' = sin 12° 30' = .2164 

71. To find the values for the cosine, tangent, and cotangent 
of a given obtuse angle A. 

Since from 67 cos A= —cos (180° — A) 

tan A = -tan (180° -A) 

and cot A = - cot (180° - A) 

Subtract the angle from 180°, find in the table the proper ratio for 
the resulting acute angle, and prefix the minus sign. 

Examples. 1. cos 150° 40'= -cos 29° 20'= - .8718 

2. tan 176° 20'= -tan 3° 40' = - .0641 

3. cot 142° 50'= -cot 37° 10'= -1.3190 

72. To find the obtuse angle A for a given sine. 

Find in the table the acute angle corresponding to the given sine 
and subtract it from 180°. 

Example. .5519 = sin 33° 30', A = 146° 30' 

73. To find the obtuse angle A for a given cosine, tangent, or 
cotangent. 

Find in the table the acute angle corresponding to tlte positive 
value of the given ratio and subtract it from 180°. 

Examples. 1. -.7254= -cos 43° 30', ,4 = 136° 30' 

2. - .5206 = -tan 27° 30', A = 152° 30' 

3. - .9601 = -cot 46° 10', A = 133° 50' 



CHAPTER VII 



THE OBLIQUE PLANE TRIANGLE 

FORMULAS FOR SOLUTION OF OBLIQUE PLANE TRIANGLES 

Formulas of the Sine 

74. Let A, B, C denote the interior angles and a, b, c the 
lengths of the opposite sides, respectively, of an oblique plane triangle. 
From A and B draw AP and B H, respectively, at right angles to 
the opposite sides, as shown. 
When the triangle is acute 
b sin C=AP=c sin B 
b c 



Hence 



sin 


B 


sin C 






a sin 


C = 


■BH = 


c sin 


A 


a 




c 






sin 


A 


sin C 




a 




b 


c 






Also 
and 



Whence 

sin A sin B sin C 

When the triangle is obtuse 

b sin C=AP=c sin B 

Hence - — - = -^— 

sin B sin C 

Also a sin C = B H = c sin (180° - A) 

, a c_ 

sin(180°-^)~^~C 

Whence 

a _ b c 

sin (180° -4) ~^TB~s^~C '^ "* ~^~ ~ c 

But by 67 sin (180° - A) = sin A 

Therefore, both for acute and obtuse triangles 
a b c 




and 



sin A sin B sin C 
a _sin A_J)_sin B_ a sin A 
b sin B c 



sin B c sin C c sin C 

Whence: The sides of a plane triangle are proportional to the 
sines of the opposite angles. 



FORMULAS FOR OBLIQUE PLANE TRIANGLES 31 



Formulas of the Cosine 

75. Let A, B, C denote the interior angles and a, b, c the 
lengths of the opposite sides, respectively, of an oblique plane 
triangle. 

From B draw BH perpendicular 
to the opposite side or to the oppo- 
site side prolonged, as shown, and let 
AH = d. 

When A is acute, from 19 

a 2 = b 2 +c 2 -2bd 
But d=c cos A 

Hence a 2 = b 2 + c 2 — 2bc cos A 

When A is obtuse, from 19 
a 2 = b 2 +c 2 +2bd 
But d=ccos(180°~,4) 

and by 67 d = c( — cos A) 

Hence a 2 = 6 2 +c 2 -26c cos A 

Therefore, in any acute or obtuse triangle 

a 2 = b 2 +c 2 -2bccos A 

Similarly b 2 = a 2 +c 2 - 2ac cos B 

c 2 = a 2 +b 2 -2abcosC 

Whence: The square of any side of an oblique plane triangle is 
equal to the sum of the squares of the other sides, minus twice the product 
of those sides into the cosine of their included angle. 

76. From the last equations 

b 2 +c 2 — a 2 




cos A = 



cos B = 



cos C = 



2bc 
a 2 +c 2 -b 2 

2ac 
a 2 +b 2 -c 2 

2ab 



Whence : The cosine of any angle of a plane triangle is equal to the 
sum of the squares of the adjacent sides, minus the square of the opposite 
side, divided by twice the product of the adjacent sides. 



32 



PLANE TRIGONOMETRY 



Formulas of the Tangent 

77. In the triangle A BC 
draw CX bisecting the angle 
C. Also draw BD and AG 
perpendicular to CX. 
Since C =180° -(A + B) 
$C = 90° -h(A+B) 

Hence 
D BC = 90° -iC=%(A+B) 
and DBF=DBC-B = \{A-B) 

Whence 



*< 




tan — - — = tan DBC = r=-= and tan 
2 DB 



Therefore 



tan|U~-S) 



(A-B) 

2 
DF 
"DC 



= tan 2)5^= 



DF 
D£ 



tan $(A+B) 
In the triangles BD.F and yl£F the angles 

BFD=AFG and BDF =90° =AGF 
Hence, by 14 DF : FG= BD : AG 

In the triangles ACG and BCD, by 14 

BD:AG = DC:GC = a:b 
Therefore DF:FG = a:b 

and DF : DF+FG = a :a+b = DF : DG 

Also, since DC : GC = a : b 

DC : DC-GC = a : a-b = DC : DG 
Hence DF(a+6) = DCXa = DC(a-6) 

DF a-6 
DC = 



and 



Therefore 



Similarly 



and 



a+b 

a-btanf(A-B) 
a+b~tanKA+B) 
a— c tan |(A— C) 
a+c~tanJ(A+C) 
b-c tan|(B-C) 



b+c tanJ(B+C) 
Whence: The difference of any two sides of a plane triangle is to 
their sum as the tangent of half the difference of the opposite angles 
is to the tangent of half their sum. 

Conventions of Algebraic Signs are not involved here because 
half the sum of any two angles of any plane triangle is less than 90°. 



FORMULAS FOR OBLIQUE PLANE TRIANGLES 33 

Formulas of the Perimeter 

78. In the triangle ABC draw AF bisecting the angle A. Also 

draw CD perpendicular to AF. B 

Then, CH= ED and by 14 AC=AD=b 

Now sin \A = —r- 

, . . 1A 2CH CD 

and 2sm|i=-r-=-r- 



Hence 4 sin 2 \A ■■ 



CD 

'' b 2 




fr T 

But by 75 (U? = b*+b*-2b> cos ^=26 2 (l-cos A) 

Therefore 4 sin 2 hA J^^-^A) 

b 2 
Whence 2 sin 2 \A = 1 — cos A 

b 2 +c 2 -a 2 



Also, from 76 cos A = 

Hence 2 sin 2 §4 = 1- 



26c 
6 2 +c 2 -a 2 2bc-b 2 -c 2 +a 2 



2bc 2bc 

^ q 2 -(6 2 -26c+c 2 ) = a 2 -(6-c) 2 
26c ~ 26c 

n ^, i a Ja-(b-c)] [a+(b-c)] (a-6+c) (a+b-c) 

Or Z' Sill o-<ri — ~ = — 

2 26c 26c 

For brevity let the perimeter a+b+c = 2s 

Then a— 6+c = 2(s — 6) and a+b — c = 2(s— c) 



Therefore 


be 


and 


sin|A- x / (s - b)(s - c) 
\ be 


Similarly 


-•sin|B-J (s - a)(s - c) 
> ac 


and 


S inK-J (S ~ a)(S " b) 



ab 

Whence, to find the sine of half an angle of a plane triangle: 
Multiply the semiperimeter minus one including side by the semiper- 
imeter minus the other including side, divide the result by the product of 
those sides, and take the square root of the quotient. 



CHAPTER VIII 



SOLUTION OF OBLIQUE PLANE TRIANGLES 

79. By the formulas established in Chapter VII, all unknown 
parts of any plane triangle may be calculated when there are known: 

I. One side and two adjacent angles 

II. One side, an angle adjacent, and the angle opposite 

III. Two sides and the angle opposite the greater 

IV. Two sides and their included angle 
V. Three sides 

80. Problem I. Given one side and two adjacent angles of a 
triangle to calculate the remaining parts. 

From 12 A + B+C = 180° 

From 74 - T ^=- r — = -A Tl 
sin A sm B sin C 

Examples. 1. Given ,4 =46° 30', 
C=31° 20', and 6 = 1000 ft., to cal- 
culate B, a, and c. 

5 = 180° -04+C) = 102° 10' 
sin^ = sin46°30' = .7254 

6 sin A 




a = 



sin B 



■■ 742.09 ft-* 



sin£ = sinl02°10' = .9775 
sin C = sin31°20' = .5200 

6 sin C 



c = 



sin B 



= 531.97 ft. 



2. Given a = 1000 f t, B = 68° 30', and C = 54° 20', to calculate 
A, 6, and c. Ans. ^ = 57° 10'; 6 = 1107.22 ft.; c = 966.80 ft. 

3. Given A = 132° 50', B = 27° 20', and c = 1000 m., to calculate 
a, 6, and c. Ans. a = 2161.21 m.; 

6 = 1353.4 m.;C = 19° 50'. 

4. To calculate the distance 
CB across a river. 

Lay off a convenient base line 
CA and measure the angles BCA 
and BAC. Let CA = 100 ft., 
BCA = 103° 10', and BAC = 71° 40'. 
Calculate CB. 

Ans. CB~ 1053.48 ft. 




SOLUTION OF OBLIQUE PLANE TRIANGLES 35 

81. Problem II. Given one side, one adjacent, and the oppo- 
site angle of a plane triangle, to calculate the remaining parts. 

From 12 .4+5+0 = 180° 

a b c 



From 74 




sin A sin B sin C 



Examples. 1. Given a = 912 ft., A = 123° 14', and C = 32° 20', 
to calculate B, b, and c. 

B = 180°-U + C)=24°26' 
sin .4 = sin 123° 14' = .8365 
sin 5 = sin 24° 26' = .4136 
sin<7=sin32°20' = .5348 

6= a|n5 = 451 ft 

sin A 

a sin C f 

c = — — 7- = 583.1 ft. 

sin A 

2. Given A = 37° 20', B = 12° 30', and b = 100 ft., to calculate C, 
a, and c. 

Ans. C = 130° 10'; a=280.2 ft; c = 353.1 ft. 

3. Given £ = 42° 30', C = 64° 20', and c = 1000 m., to calculate 
A y a, and b. 

Ans. ^ = 73° 10'; a = 1062.02 m.; 6 = 749.59 m. 

4. Given A = 147° 23', C = 16° 19', and c = 1000 ft., to calculate 
B, a, and b. 

Ans. 5 = 16° 18'; a = 1918.8 ft.; 6 = 999.2 ft. 

5. To calculate the distance to the moon. 




■■© 



Let E be the center of the earth, P a point on its surface, and 
let M be the center of the moon. When the angle EPM = 105° 30', 
the angle EMP = 55'. The distance EP = 3960 miles. Calcu- 
late EM. Ans. EM = 238,500 miles, nearly 



36 



PLANE TRIGONOMETRY 



82. Problem III. Given two sides and the angle opposite the 
greater given side of a plane triangle, to calculate the remaining parts. 



From 12 A+B+C = 180° 



From 74 



sin A sin B sin C 




Examples. 1 . Given b = 1600 ft., c = 2000 ft., and C = 47° 20', 
to calculate A, B, and a. 

sinC = sin47°20' = .7353 

s i n 5 = ^ m _^ = -588 3 = s i n 36 o 02 > 
c 

^ = 180°-CB+(7) = 96 o 38' 

sin ^4 = .9933 

6 sin A 



a = 



sin £ 



= 2701.5 ft. 



2. Given o = 1000 m., b =2000 m., and 5 = 132° 10', to calculate 
A, C, andc. Ans. ^ = 21° 45'; C=26° 05'; c = 1186.5 ft. 

3. Given b = 1000 ft., c = 100 ft., and £ = 49° 20', to calculate A, 
C, and a. Ans. ^4 = 126° 19'; C=4° 21'; a= 1062.3 ft. 

4. Given a = 1060.43 m., c = 1000 m., and A = 62° 19', to calcu- 
late B, C, and b. Ans. £ = 61° 04'; C=56° 37'; 6 = 1048.1 m. 

5. To calculate the distance across a lake from direct measure- 
ments of AB, BC, and ACB. 




Let ,45 = 2000 ft., 5C = 1000 ft., and ACB = 41° 53'. Calcu- 
late AC. Ans. 2629.7 ft. 



SOLUTION OF OBLIQUE PLANE TRIANGLES 37 

83. The Anomalous Case of Problem III, Page 36. Should 
two sides and the angle opposite the shorter side be given, the 
remaining parts of an oblique triangle are theoretically indeter- 
minate. This cannot happen, however, when the given angle is 
obtuse for, by 13, that angle must be opposite the greater given side. 

When Given Angle Is Acute. b 

In the oblique triangle ABC let A, 
a, and c be given and let a be less 
than c. 

Draw B H perpendicular to AC, 
lay off HD=HC, and draw BD, 
which, by 16, is equal to a. 

In the triangles ABC and A BD ^~ 
the given parts A, a, c, are the same, while the unknown parts are 
entirely different. 

Example. To calculate the distance across a lake from direct 
measurements of AB, BC, and BAC. 





Let ^5 = 2000 ft., £(7 = 1000 ft., and BAC=l§° 30'. Calcu- 
late AC. 

Since BC is less than AB, the line BD might be taken equal to 
BC and the angle at B, which determines the required distance, might 
be either ABC = 180° -(B AC + AC B) 

or ABD = 180°-(BAC+ADB) 

Thus the calculation, instead of giving the correct distance AC — 
2629.7 feet as in 82, example 5, might give the incorrect distance 
AD = 1140.8 feet. 

Therefore, care should be taken to measure the angle opposite 
the greater given side unless circumstances indicate which result 
to accept. 



38 



PLANE TRIGONOMETRY 




84. Problem IV, Method A. Given two sides and the included 
angle of a plane triangle, to calculate the remaining parts. 

From 75, by formulas of the J 
cosine 

a 2 =6 2 +c 2 -26c cos A 
b 2 = a 2 +c 2 -2ac cos B 
c 2 = a 2 +6 2 -2a6 cos C 

Examples. 1 . Given A = 150° 22', 6 = 720 ft., and c = 900 ft., to 
calculate a, B, and C. 

a 2 = b 2 +c 2 -2bc cos A = 2,454,883 
o = 1566.81 ft. 
a2+c 2_fr2 a 2 +6 2 -c 2 ft . CQ 

cos £ = — = .9738 cos C = — — -r — = .9588 

2ac 2ab 

Whence 5 = 13° 08' and C = 16°30' 

2. Given A = 18° 39', b = 100 m., and c = 200 m., to calculate a, 

B, and C. Ans, o = 110 m.; 5 = 16° 54'; C = 144° 27' 

3. Given B = 65° 20', a = 300 ft., and c = 400 ft., to calculate A, 

C, and 6. Ans. ^4 = 44° 46'; C = 69° 54'; 6 = 387.1 ft. 

4. Given C=131° 21', a =250 m., and 6 = 1000 m., to calculate 
A,B,&ndc. Ans. ^ = 9° 9'; B = 39° 30'; c = 1180.2 m. 

5. Given A = 101° 42', 6 = 200 ft., c = 100 ft., to calculate a, B, 
andC. Ans. a= 241.06 ft. ; 5 = 54° 20'; C = 23° 58' 

6. To calculate the distance and direction of an objective point 
that cannot be seen from the initial point. 




Let A be the initial point and B the objective point. Given 
^C=1000 ft., BC =700 ft., and ^5 = 110° 36', to calculate AB 
and CAB. Ans. ^.B = 1408.02 ft. ;(L4J3= 27° 44' 



SOLUTION OF OBLIQUE PLANE TRIANGLES 39 



85. Problem IV, Method B. Given two sides and the included 
angle of a plane triangle, to calculate the remaining parts. 
From 77, by formulas of the tangent 

tan \(A— B) a—b 

tan |U +B)~a+b 

tan j(A — C) a—c 




tan |U+C) a+c 
tan |(.g-C) b-c 
tan %(B+C)~b+c 

Examples. 1. Given a = 1000 ft., 6 = 200 ft., and C = 72° 40'. 
to calculate A, B, and c. 

a+6 = 1200 ft. a- 6 = 800 ft. 
By 12 K^+£)=2(180 o -C) = 53 o 40' 

tan i(A+B)=t&n 53° 40' = 1.3597 

tan \{A - B) = ^ tan \{A + B) 
a-\-o 

SDft ft 
or tan %(A-B) = ^^-X 1.3597 = .9065 = tan 42° 11' 

lZUU It. 

^4+|£ = 53°40' 



Adding 
Subtracting 

By 74 



c = - 



\A-\B=42° 11' 

^ = 95° 51' 

J5 = ll°29' 

a sin C b sin C 



= 959.0 ft. 



sin A sin B 

Since all the angles are known, the side c might have been calcu- 
lated by either one of the other 
formulas of the tangent. The for- 
mula of the sine is, however, more 
convenient. 

2. To calculate the distance 
and direction of a point that cannot 
be seen from the initial point. 

Suppose A is the initial point 
and B the point sought. Given ^ c 

,40 = 400 ft., C£ = 1000 ft., and 4CB = 92° 34', to calculate AB 
and CAB. Ans. 45 = 1094.0 ft.; 045=66° 00' 




40 



PLANE TRIGONOMETRY 



86. Problem V, Method A. Given the three sides of a plane 
triangle, to calculate the angles. 

From 75, by formulas of the cosine 
6 2 +c 2 -a 2 



cos A = 
cos 2? = 
cos C= 



26c 

a 2_|_ c 2_&2 

2ac 

a 2 +6 2 -c 2 

2ab 




Examples. 1. Given a = 236 ft., 6 = 312 ft., and c = 480 ft., to 
calculate A, B, and C. 

a 2 = 55,696 6 2 = 97,344 c 2 = 230,400 

2a6 = 147,264 2ac = 226,560 26c = 299,520 

272,048 D 188,752 _ -77,360 



C0S ^ = 299,520 
cos 4 = .9083 
4=24° 44' 



COS5 = 226,560 
cos B = . 8331 
2? = 33° 35' 



C0S C= 147,264 
cosC= -.5253 
C= 121° 41' 



2. Given o=4 m., 6 = 6 m., c=8 m., to calculate A, B, and C. 

Ans. ^ = 28° 57'; 5=46° 34'; C = 104° 29' 

3. Given a =1000 ft., 6=700 ft., c=500 ft., to calculate A, B, 
andC. Ans. 4 = 111° 48'; B = 40° 32'; C=27° 40' 

4. In a steam engine the length of the crank arm SK is 18 
inches and the length of the connecting rod KG is 7 feet 6 inches. 




Calculate the positions of the crank and connecting rod when the 
engine is at one-quarter stroke from the forward center, at which 
point 

SC=SK+ KC-\SK= KC+ISK 

Therefore SC = 8 ft. 3 in. 

cos CSK = .5682 cos CKS = - .4250 cos KCS = .9864 

CSK = 55° 23' CKS = 115° 09' KCS - 9° 28' 



SOLUTION OF OBLIQUE PLANE TRIANGLES 41 



87. Problem V, Method B. Given the three sides of a plane 
triangle, to calculate the three angles. 

From 78, by formulas of the perimeter 



sin 




-r- 


-Ms- 
be. 


-c) 


sin 


-V<" 


-a)(s- 
ac 


-c) 


sin 


■V* 


-a)(s- 
ab 


-b) 




Examples. 1. Given a = 126 ft., 6 = 374 ft., and c = 452 ft., to 
calculate A, B, and C. 

a=126ft. s-a = 350ft. 
6 = 374 ft. s-6 = 102 ft. 
c = 452ft. s-c= 24 ft. 



2s = 952 ft 
sin \A 

sin \B 



-i 

-A 

sin %C=^ 



102X24 
374X452 
350X24 



s = 476 ft. 



= .1203 = sin6°55' 



126X452 
350X102 



= .3840 = sin22°35' 



.8704 = sin60°30' 



126X374 

Whence ^ = 13° 50' 5 = 45° 10' (7 = 121° 00' 
,4 + 5+0=180° 00' 

2. Given a = 200 ft., 6 = 300 ft., c = 400 ft., to calculate A, B, 
andC. Ans. ,4 = 28° 58'; 5 = 46° 34'; C= 104° 28' 

3. In a steam engine the length of the crank arm SK is 2 feet 
and the length of the connecting rod KC is 9 feet. Calculate the 




positions of the crank and connecting rod when the engine is at 
one-quarter stroke from the forward center, at which point 
SC=10feet. 

Ans. CSK= 54°. 54'; CKS = 114° 38'; KCS = 10° 28' 



42 



PLANE TRIGONOMETRY 



MISCELLANEOUS EXAMPLES IN OBLIQUE TRIANGLES 

1. A and C are two 
stations and B is a buoy. 
Given AC = 100 ft., £^C = 
85° 20', and £(7,4 = 82° 40', 
to calculate AB. 

Ans 45 = 477 ft. 

2. In running a line X'X 
a surveyor, in order to avoid 
a building, turned 20° to the 
right at A and measured 200 
feet to C. At C he turned 
40° to the left and measured 
CB to a point B on his origi- 
nal line. Calculate CB, AB, 
and CBX. 

Ans. CB = 200 ft.; ,45 = 375.9 ft.; C£X=160° 




3. The distance between two points 
A and B on the deck of a ship is 100 feet. 
From each of these points the masthead 
C of another ship was observed. Given 
B AC = 88° 45' and ABC = 90° 32', to 
calculate BC. 

Ans. £C = 7998ft. 



4. In the triangle ABC, 
A = 74° 13', a = 1000 ft., and 
c = 544 ft. Calculate B, C, 
and 6. 




Ans. 5 = 74° 13'; C = 31° 34'; 6 = 1000 ft. 



SOLUTION OF OBLIQUE PLANE TRIANGLES 43 



5. In order to avoid a 
swamp, a surveyor turned 
11° 01' to the right at a point 
A and measured 1000 feet to 
C. He then turned 83° 51' 
to the left and ran a line to a 
point B on his original line. 
Calculate CB, AB, and ABC. 

Ans. C5 = 200 ft.; ,45 = 1040.5 ft. 




ABC = 72° 50 r 




6. In order to establish a railroad line between points A and B 
so that a tunnel might be 
driven from both ends, an 
engineer selected a point C 
from which both A and B 
could be seen. He then found 
by direct measurement AC = 
1000 ft., AC B = 95° 48', and 
CB = 809.5 ft. Required the * 
angles BAG and ^4J5C and the distance AB. 

Ans. BAC = ZQ° 40'; ABC = ±7° 32'; ^£ = 1348.7 ft. 

7. The distance SM of the planet Mars from the sun is about 
1.52 times the distance of the earth from that luminary. The angle 
SEM between the lines from 
the earth to the sun and to 
Mars was observed to be 
147° 20', and the distance ES 
was assumed to be 93 million 
miles. Calculate the follow- 
ing: 

(1) The angle SME subtended at Mars by a line from the 
earth to the sun. 

Ans. 20° 48' 

(2) The angle ESM subtended at the sun by a line from the 
earth to Mars. 

Ans. 11° 52 

(3) The distance EM of Mars from the earth. 

Ans. 53,850,000 miles, about 




44 



PLANE TRIGONOMETRY 



9 
A 



/ 



/ 



/ 



£ 



H 



8. From a ship sailing due north 
at the rate of 12 miles per hour, a light- 
house bears N. 37° 42' E. Ten minutes 
later it bears N. 62° 34' E. Calculate ./ 
the distance of the lighthouse from the 
last point of observation. 

Ans. 2.9 miles 

9. The sides of a triangle ABC are 
a = 7 m., 6 = 8 m., c = 9 m. Calculate the 
angles. 
Ans. ,4 =48° 11'; 5 = 58° 25'; (7 = 73° 24' 

10. A surveyor, in order to avoid a thick wood, started from A 
and measured AC, ACB, 
and CB. He then calcu- 
lated AB and the direction 
he should take to prolong 
the line AB. Given AC= 
1000 ft., ^C5 = 164° 20', 
and C£ = 2000 ft., to calcu- 
late AB and the angle to 
be turned from BC so as 

to continue the line AB. Ans. AB = 2975.1 ft.; C5X= 174° 48' 

11. To determine the distance between two inaccessible points 
A and C, a surveyor 
selected a point D in line 
with A and C, turned off 
an angle ADB, and meas- 
ured DB, DBA, and 
DBC. Given DB = 1000 
ft., ADB = 5±° 30', DBA 
= 47° 30', and £5(7 = 87° 
50', to calculate AC. 

From the triangle DBC calculate DC. Ans. 
From the triangle DBA calculate DA. Ans. 
Whence ,4(7 = 881.5 ft. 

12. The angles of a triangle ABC are A = 132° 43 7 , 5 = 21° 36', 
and C = 25° 41', and the shortest side is 1000 feet. Calculate the 
other sides. Ans. a = 1995.9 ft.; c = 1177.4 ft. 





DC =1635.3 ft. 
DA = 753.8 ft. 



SOLUTION OF OBLIQUE PLANE TRIANGLES 45 




Calculate CB. 



13. To determine 
the distance between two 
buoys C and B, the fol- 
lowing direct measure- 
ments were made : AD = 
500 ft., BAC=\4° 30', 
DAC = 42>° 10', ADC= 
85° 40', and £1X7=16° 

From triangle ADC, ^4C=640 ft. and DC=439.1 ft. 
From triangle ADB, .45 = 1407.5 ft. and DB = 1215.8 ft. 
Whence, from triangle ABC, and also from triangle DBC, 

CB= 804.0 ft. 

14. The sides of a triangle ABC are a = 3 m., b = 5 m., c = 7 m. 
Calculate the angles. Ans. 4=21° 47'; 5 = 38° 13'; C = 120° 00' 

15. To calculate 
the distance between two 
points B and F, a sur- 
veyor made direct meas- 
urements on the opposite 
side of a river, as follows: 
AC =400 ft., F=80°33', 
P = 30° 18', Q=29° 23', 
and Z= 94° 47'. Calcu- 
late BF. 

From triangle ABC, S 
584.4 ft. 

From triangle AFC, T=25° 32', AF =767.9 ft., and CF= 468.2 ft. 
Whence, from triangle ABF and also from triangle CBF, 

BF= 778.7 ft. 




= 39° 46', AB = 306.8 ft., and CB = 



CHAPTER IX 



tan A = : 



MISCELLANEOUS PROPERTIES AND RELATIONS OF 
PLANE TRIANGLES 

88. In any oblique plane triangle with sides and angles as 
shown, draw BH perpendicular to AC 
When the triangle is acute 
BH 
AH 
But BH = a sin C 

and AH = b — a cos C 

Therefore tan A = z ^ 

b — a cos C 

When the triangle is obtuse 

BH 

AH 
But BH = a sin C 

and AH = a cos C—b 



tan (180°-^)= - 



Hence 



tan(180°~^l) = 




a cos C—b b — acosC 
Now by 67 tan (180° - A) = - tan A 

Whence, both for acute and obtuse triangles 

a sin C 



Similarly 
and 



tan A = 
tanB = 
tan C = 



b—a cos C 
b sin A 

c—b cos A 
c sin B 

a— c cos B 



Similarly, it may be shown that 



and 



cot A = 
cotB = 
cotC= 



b—a cos C 
a sin C 

c—b cos A 
b sin A 

a— c cos B 
c sin B 



PROPERTIES AND RELATIONS OF TRIANGLES 47 

89. In any plane triangle, from 32 and 78 

cos 2 ^ = l-sin 2 ^ and sin 2 i A J s ' h ^ s - c ) 

tt »ia i (*-b)(s-c) (bs+cs-s 2 ) s(s-a) 
Hence cos 2 \A = \ r — 7 = — 7 



Whence cos iA = J 



be 



r.. -i i ir» /s(S-b) , ,^ /S(S-C) 

Similarly cos |B = ^ and cos |C = ^/ — - — 

90. In any plane triangle ABC, from 78 and 89 

sin 2 i A = = and cos 2 J^l = — =- 

6c be 



Whence, by 33 tan|A = J (s b)(s c) 

\ s(s— a) 



o- -ii , in /(s-a)(s-c) , . „ /(s-a)(s-b) 
Similarly tan |B = ^ -^^p and tan |C=^ -^z^" 

91. In any plane triangle ^12?C, draw AF bisecting the angle 
A. Also draw BD perpendicular to AF. 

Then in triangle ABB, by 74 

csin^ = 5Z)sin^DB J* 

by 14 AD = c, HD=HB 

Hence BD = 2c sin \A 
Now ADB = 90°-$ A 

and sin ^42) JS = cos \A 

Therefore 

sin ^4 = 2 sin \A cos ^ /f *~ zT 

Now substitute in this last equation from 78 




sin 2 



M = J('-6)(*-c) and from 89 cosM = J*zfO 
\ 6c > oc 



Whence sin A= 2Vs(s-a)(s-b)(s-c) 

be 



Similarly sin B = 2Vs(s-a)(s-b)(s-c) 

ac 

and sin c= 2Vs(s-a)(s-bKs-c) 

ab 



'48 



PLANE TRIGONOMETRY 



92. In any oblique plane triangle with sides and angles as 
shown, draw AH perpendicular b 

to BC. 

When B and C are acute 
a = d+e 
But d = b cos C 

and e = c cos B 

Hence a = b cos C+c cos 2? 

When B is obtuse 

a = d— e 
But d = b cos C 

and e = c cos (180° -5) 

or by 67 0= — c cos 5 

Hence a = b cos C+c cos J5 




In each case 
Similarly 
and 



a = b cos C+c cos B 
b=ccos A+a cos C 
c = a cos B+b cos A 



S1T1 A ft 

93. From 74 - — 5 = -7- in any plane triangle ABC 



Hence 



sin A 



H-i 



and 



sin B 
sin A — sin B a—b 



, sin .4 a 

and - — = + 1=^+1 
sin B 

sin^4 + sinJ5 a+b 



sin B 
Whence by division 



b 

sin A 



and 



sin B 
sin B a—b 



But from 77 



Therefore 



Similarly 



and 



sin ^4+sin B a+b 

tan %(A — B) _a— b 
tan %(A+B)~ a+b 

sin A— sin Btan §(A— B) 
sin A+sin B~tan |(A+B) 

sin A — sin C tan ^(A— C) 
sin A+sin C tan |(A+C) 

sin B-sin C tan ^B-C) 
sin B+sin C"tan |(B+C) 



PROPERTIES AND RELATIONS OF TRIANGLES 49 



94. In any plane triangle^ 
ABC, by 92 

a = b cos C+c cos B 
Dividing through by a 

1 = — cos C-\ — cos B 
a a 



Now by 74 



Therefore 



a 

1 = 



sin B 
sin A 
sin B 




and — = . 

a smA 

COS C-\ — ; 7 COS B 

sin A 



sin ^4 
Whence, clearing of fractions 

sin A = sin B cos C+cos B sin C 
Similarly sin B = sin A cos C+cos A sin C 

and sin C=sin A cos B+cos A sin B 

95. In any plane triangle ABC, by 76 



cos A- 



b 2 +c 2 



2bc 



Draw AH perpendicular to BC. 
Then by 18 

and c 2 = e 2 +f 2 

By construction 
a = d+e 
Squaring both sides 

a 2 = d 2 +2de+e 2 
Fromr these values of b, c, and a 
d 2 +f 2 +e 2 +f 2 - 




cos A = - 



■d 2 -2de-e 2 f 2 -de 



But 



^- = sin£, 
c 



2bc 
de f f d e 
be b c b c 

f • e 

— = sin C, — = cos B y 
c 



be 



t 

be 



and -r- = cos C 
b 



Whence, by substitution 

cos A = sin B sin C— cos B cos C 
Similarly cos B = sin A sin C— cos A cos C 

and cos C = sin A sin B — cos A cos B 



CHAPTER X 



CALCULATION OF AREAS 

AREA OF PLANE TRIANGLE 

96. Given two sides and their included angle of a plane tri- 
angle ABC, to calculate its area. 

Let T denote the area and draw BH perpendicular to AC. 

3 



Then by Geometry 






r= 


-hbxBH 


Now 


BH= 


= c sin A 


Therefore 


T= 


= |bc sin A 


Similarly 


T= 


= §ac sin B 


and 


T= 


= §ab sin C 




a b £ 

Whence, the area of a plane triangle is equal to one-half the 
product of any two of its sides multiplied by the sine of their included 
angle. 

Examples. 1. In a plane triangle given A = 142° 32', 6 = 200 ft., 
and c = 300 ft. Calculate the area T. 

T=\bc sin A = \ 200 ft. X 300 ft.X.6083 
Whence 7=18,249 sq. ft. 

Ans. Area = 18,249 sq.ft. 

2. In surveying a trian- >a 
gular tract of land, the follow- 
ing direct measurements were 
made: AB = 10 chains, AC = 
11 chains, and BAC = 41° 18'. 
Calculate the area (10 square 
chains equal 1 acre). 

Ans. 3.63 acres 

m c 

3. In a plane triangle given -45 = 50 chains, AC = 40 chains, 
and BAC = 39° 3'. Calculate the area. Ans. 63.00 acres 




CALCULATION OF AREAS 



51 



97. Given the three sides a, 6, c, of a plane triangle, to cal- 
culate its area T. 

T = \bc sin A 



From 96 
From 91 

Whence 
Examples. 



1. 



sin A = 2y/s(s—a){8~-b)(s — c) 
be 

T = Vs(s--a)(s-b)(s-c) 
In survey- 



ing a triangular tract of land, a 
surveyor made the following 
direct measurements: a=20 
chains, 6 = 50 chains, and c = 60 
chains. Calculate the area. 

Ans. 46.837 acres 
2. Given a = 30 chains, 6 = 40 chains, and c = 50 chains. Cal- 
culate the area. Ans. 60.00 acres 




AREA OF REGULAR POLYGON 

98. Let A be the cen- 
ter and B and C adjacent 
vertices of a regular poly- 
gon. Now, let n= number 
of sides, r= radius of cir- 
cumscribed circle, and S = 
area of polygon. Draw AB 
and AC. 
Then, by Geometry 

AB = AC = r, S=nXarea ABC, and angle BAC = 




360° 
n 



From 96 



area ABC = \r* sin BAC=hr 2 sin 



360° 



n 



Whence 



S = *r 2 nsin 



360° 



Examples. 1. Given n = 10 and r = 10 ft. Calculate area of 
polygon. 

S = i X 10 2 XlO sin 36° = 293.90 sq. ft. 
2. Given n = 36 and r = 1 ft. Calculate area of polygon. 
S = |X1 2 X36 sin 10° = 3.125 sq. ft. 



CHAPTER XI 




RELATIONS OF RATIOS FOR VARIOUS ACUTE ANGLES 

RATIOS FOR SUM OF ANY TWO ACUTE ANGLES 

99. Let VAX=X and ^ 
VAY=Y be any two acute 
angles. 

Then YAX=X+Y 

From any point B on AY 
draw BC perpendicular to AV, 
forming the oblique triangle 
ABC. 
Then, from 94 

sin ^4 = sin B cos C+cos B sin C 
But A=X+Y, £ = 90°-F, andC = 90°-X 

Hence, by identity sin A = sin (X+ Y) 
and from 38 sin B = cos Y cos B = sin Y 

sin C=cos X cos C = sin X 

Whence sin (X+Y) =sin X cos Y+cos X sin Y 

100. In the plane triangle ABC, shown in 99, we have, from 95 

cos A = sin B sin C— cos B cos C 
Whence, from the relations given for the sine in 99 
cos (X+Y) =cos X cos Y-sin X sin Y 

101. From 33 and from the relations just given for sine and 
cosine 



tan(Z+7) = 



sin (Z+ Y) sin X cos F+cos X sin Y 



cos (Z+ Y) cos X cos Y — sin X sin Y 
Dividing numerator and denominator by cos X cos Y 

sin X cos Y , cos X sin Y 



tan(Z+y) = 



Whence, by 33 



tan (X+Y) 



cos X cos Y cos X cos Y 
cos X cos Y sin X sin Y 
cos X cos F cos X cos F 

tan X+tan Y 



1-tan X tanY 



RELATIONS OF RATIOS FOR ACUTE ANGLES 53 

102. From 34 and from relations given in 99 and 100 

Y ^ cos (X+Y) _ cos X cos Y— sin X sin F 
cot {X + Y)- gin (Z+ y) - gin z cog y+cos z gin Y 

Dividing numerator and denominator by sin X sin Y 

cos X cos F sin X sin F 

, / xr , T7\ sin X sin Y sin X sin F 

COt (X+F)= — r 



sin X cos F cos X sin F 
sin X sin F sin X sin F 
cotXcotY-1 



Whence, by 34 cot (X+Y)- ^ Y+cot x 

103. Formulas in 99, 100, 101 and 102 being true for any acute 
angles X and Fare true when their sum is greater than 90°. 

RATIOS FOR DIFFERENCE BETWEEN TWO ACUTE ANGLES 

104. Let angle VAX=X 
and angle VAY=Y. 

Then YAX=X-Y 
Through any point B on AY 
draw HBC perpendicular to 
AV forming the oblique tri- 
angle ABC. 
Then, from 94 g 

sin A = sin B cos C+cos B sin C 
But A=X-Y, 5 = 180°- (90° -F), andC = 90°-X 

Hence, by identity sin A = sin ( X— Y) 
and by 38 and 67 sin 5 = cos F cos 5= —sin F 
sin (7= cos X cos C=sin X 
Whence sin (X— Y) =sin X cos Y— cos X sin Y 

105. In the plane triangle ABC shown in 104, we have, from 95 

cos A = sin B sin C— cos B cos C 
Whence, from the relations given in 104 

cos (X— Y) =cos X cos Y+sin X sin Y 

106. From 33 9 104, and 105 and by the process shown in 101 

tanX-tanY 




tan(X-Y)=- 
l» 104, and 10! 

cot(X-Y)=* 



1+tan X tanY 
107. From 34, 104, and 105 and by the process shown in 102 

cot XcotY+1 
cot X -cot Y 



54 PLANE TRIGONOMETRY 

RATIOS FOR SUM AND DIFFERENCE OF TWO ACUTE ANGLES 

108. From 99 and 104, by addition and subtraction 

sin (X+Y)+sin (X- Y) =2 sin X cos Y 
and sin (X+Y) -sin (X- Y) =2 cos X sin Y 

Similarly, from 100 and 105 

cos (X+Y) +cos (X- Y) = 2 cos X cos Y 
and cos (X+Y) -cos (X- Y) = -2 sin X sin Y 

RATIOS FOR TWICE AN ACUTE ANGLE 

109. The formulas given in 99, 100, 101, and 102 being true for 
any two acute angles, are true if Y=X. 

Making F=X in each case, there results 

From 99 sin 2X=2 sin X cos X 

From 100 cos 2X = cos 2 X-sin 2 X 

and by 32 cos 2X= 1 -2 sin 2 X=2 cos 2 X- 1 

2tanX 



From 101 tan2X = 

From 102 cot2X= 



l-tan 2 X 

cot 2 X-l 

2cotX 



RATIOS FOR HALF AN ANGLE 

110. The formulas in 109, being true for any acute angle X, 
are also true for %X. 

Therefore sin X = 2 sin |X cos \X 

Whence sin JX = t^> and cos |X = — — : — — 

2 cos |X ^2 sin |X 

From 109 cos 2X=cos 2 X-sin 2 X 

Changing X to \ X and transposing, there results 

cos 2 \X— sin 2 |X=cos X 

From 32 cos 2 £X+sin 2 \X = 1 

By subtraction 2 sin 2 \X= 1 — cos X 

By addition 2 cos 2 \X = 1 +cos X 



Whence sin §X = ^ l ~ cosX a nd cos §X = yj 1+C0S X 

From 33 and 34 tan £X = \/ ~ C ° S ^ and cot £X = \/ — 

\ 1+cos X M- 



+cos X 



cos X 



RELATIONS OF RATIOS FOR ACUTE ANGLES 55 

RATIOS FOR WHOLE AND HALF ANGLES 

111. In formulas in 108 make X+ Y = A and X- F = 5. 
Then X=$(A+B), Y=i(A-B) 

Mid sin A+sin B= 2 sin KA+B) cos i(A-B) 

sin A-sin B = 2 cos §(A+B) sin |(A-B) 
cos A + cos B= 2 cos J(A+B) cos J(A-B) 
cos A- cos B= -2 sin |(A+B) sin £(A-B) 

Dividing and applying 33 and 34 

sin A — sin BtanJ^A— B) , cos A — cos B_ tan £(A— B) 
sin A+sin B~tan §(A+B) an cos A+cos B~ ~~~cot KA+B) 

EXAMPLES FOR PRACTICE 

1 12. Prove the following relations : 

1. sin (x+y) cos y — cos (x+y) sin y = sin x 

2. cos (x+y) cos (x— y)=cos 2 a; — sin 2 y 

tan 2 x— tan 2 i/ 



3. tan (x+y) tan (x—y) = 

4. cot (x+y) cot (x—y) 



1 — tan 2 x tan 2 y 
cot 2 x cot 2 y — 1 



cot 2 y — cot 2 x 

5. sin 2# = (sin #+cos a:) 2 — 1 

6. cos 2x = cos 4 x — sin 4 x 7. cot x — tan a; = 2 cot 2z 

sinar+siny w , \ ft sin (x+y) tanar+tany 

8. ■ - = tan|(a;+y) 9. ? ^ = T~TT I - ^ 

cosa;+cosy cos (x—y) 1+tanxtany 

--- . n 2tanx , 1 — tan 2 a; 

10 - ""^-i+ssi and cos2x= t+^ 

ii. tan a,- 2 "** 12. ^fe^= ^ na; 7! any 

cot 2 x— 1 Sln 0&+y) tan x+tan y 

10 sin x— sin y M t/ , % 

13. £= _ cot J(z+y) 

cos x— cosy 

1 , sin 2x 

14. tanx = : 



1+ cos 2a; 



15. sin x+cos x= = t Vl+sin2x 



16. cos x— sin x= =*= y/\ — sin 2x 

17 , , sin a; , 1 sin a; 

17. tan J a; = — — and cot %x = 



1+cosa; 1 — cos a: 



56 PLANE TRIGONOMETRY 

RATIOS FOR MULTIPLE ANGLES 

113. From formulas given in 108 

sin (Z+ F)+sin (Z- F) =2 sin Z cos F 
or sin (Z+ F) = 2 sin Z cos F-sin (X- 7) 

Now make X=(n—l)x and F = a: 

then sin nx=2 sin (n— l)x cos x— sin (n— 2)x 

Making n — 2 sin 2a: = 2 sin a: cos x 

n = 3 sin 3# = 2 sin 2x cos a:— sin x 

n=4: sin 4a: = 2 sin Sx cos a:— sin 2a: 

n=5 sin 5a: =2 sin 4a: cos a:— sin 3a:, and so on 

114. Again, from formulas given in 108 

cos (Z+ F)+cos (Z- F) = 2 cos Z cos F 
or cos (Z+ F) = 2 cos Z cos F- cos (Z- F) 

Now make X=(n—l)x and F = x 

Then cos nx=2 cos (n— l)x cos x— cos (n— 2)x 

Making n = 2 cos 2.t = 2 cos 2 x — 1 

n=3 cos 3z = 2 cos 2a: cos a:— cos x 

n=4: cos 4r = 2 cos 3x cos a:— cos 2a: 

n=5 cos 5a: = 2 cos 4a: cos x— cos 3a:, and so on 

115. The sine and cosine of a multiple angle being known, the 
tangent and cotangent may be determined from the formulas 

sin nx , cos nx 

tan nx = and cot nx = ~ 

cos nx sin nx 

116. To find the ratios for 3a: in terms of the ratios for x. 

From 1 13 sin 3a: = 2 sin 2a: cos a:— sin a: 

and sin 2a: = 2 sin x cos x 

Therefore sin 3a: = 4 sin x cos 2 a:— sin a: 

and from 32 sin 3a: = 4 sin x (1 — sin 2 a:) — sin x 

Whence sin 3x=3 sin x— 4 sin 3 x 

Similarly, from 114 cos 3x=4 cos 3 x— 3 cos x 

3 tan x— tan 3 x 



From 33 tan3x= 

From 34 cot3x = 



l-3tan 2 x 

cot 3 x— 3 cot x 

3 cot 2 x- 1 




RELATIONS OF RATIOS FOR ACUTE ANGLES 57 

1 17. Let the acute angle 
Y AX be divided into n equal 
parts, each part equal to 2x, 
as shown. 

Then YAX = 2nx 

With A as a center and r 
as a radius, describe an arc of 
a circle intersecting A Y at B 
and AX at C. Draw CE per- 
pendicular to AX, B E perpen- 
dicular to AY, and draw £C Construct BFGC as a portion of the 
regular inscribed polygon of the circle and BKLM€ as a portion of 
the regular circumscribed polygon, each with sides subtending at 
the center an angle equal to 2x, as shown. 

Then by Geometry and for any value of n as a whole number 
CE=BE and CE+BE=2CE 
Also BC<BF+FG+GC 

BF+FG+GC<BK+KL+LM+MC 
and BK+KL+LM+MC<2CE 

Now BC = 2r sin nx 

BF+FG+GC = 2rnsmx 
BK+ KL+LM+MC=2m tan x 
and 2CE = 2r tan nz 

Therefore 

2r sin nx<2rn sin z<2m tan x<2r tan na: 
Dividing through by 2r there results 

sin nx<n sin #<?i tan #<tan nx 
Now let m by any number whatever 

^, sin nx n sin x n tan # tan na; 

Then < < < 

m m m m 

These relations being true for any acute angle divided into 2n 

90° 
equal parts are true for any integral value of n less than — . 

LtX 

They are evidently also true for any integral or fractional value of m. 
The results shown are of great importance in the calculation of 
tables of Natural Sines, Cosines, Tangents, and Cotangents. 



58 



PLANE TRIGONOMETRY 



APPROXIMATE RELATIONS OF RATIOS FOR SMALL ANGLES 

118. Let any small angle A be divided 
into n equal parts, each equal to x. 

Then A=nx 




From 117 
Hence 



From 33 

Therefore 



sin nx<n sin x<tan nx 
n sin a: — sin nx<tan nx— sin nx 
sin nx 



- = tan nx 



cosnx 

sin nx = tan nx cos nx 
and tan nx — sin nx = tan nx — tan nx cos nx 

or tan nx — sin no; = tan nx(l — cos nx) 

Hence n sin x — sin nx < tan nx(l — cos nx) 

From 30, as nx diminishes tan nx diminishes while cos nx increases. 
Consequently, tan nx (1 — cos nx), tan nx— sin nx, and n sin x— sin nx 
continually diminish as nx diminishes. 

Therefore, if nx is so small that tan nx— sin nx may be neglected 
in practice, then n sin x— sin nx may also be neglected in practice 
and when nx = 

sin nx = n sin x = tan nx 
Whence, if tan nx = sin nx+ a negligible value, then, for practical 
purposes, it may be assumed that 

n sin x = sin nx 
sinnx 



or, that 



sin x = 



n 



119. By 117 sin nx<n tan x<tan nx 
Therefore, if nx is so small that tan nx — sin nx may be neglected 
in practice, then tan nx—n tan x may also be neglected in practice and 
it may be assumed that 

n tan x = tan nx 
tannx 



or, that 



tan x = - 



n 



CHAPTER XII 

CALCULATION OF RATIOS 
CALCULATION OF PARTICULAR RATIOS 

120. Ratios for 45°. 

From 32 sin 2 45°+cos 2 45° = 1 

and from 38 sin 45° = cos (90° - 45°) = cos 45° 

Hence sin 45° = cos 45° 

Therefore 2 sin 2 45° = 1 

and 2 cos 2 45° = 1 

Whence sin 45° =* § y/2 = .707107 

and cos 45° = § \/2 = .707107 

Also, from 33 tan 45° = 1 = 1 .000000 

and from 34 cot 45° = 1 = 1 .000000 

121. Ratios for 30°. 

From 109 2 sin x cos z = sin 2x 

Therefore 2 sin 30° cos 30° = sin 60° 

But by 38 sin 60° = cos (90° - 60°) 

or sin 60° = cos 30° 

Hence 2 sin 30° cos 30° = cos 30° 

and 2 sin 30° = 1 

Therefore sin 30° = \ = .500000 

and from 32 cos 30° = \ y/z = .866025 

Also, from 33 tan 30° = i \/3 = .577350 

and from 34 cot 30° = \/3 = 1.732051 

122. Ratios for 15°. 

From 104 sin 15° = sin 45° cos 30°-cos 45° sin 30° 
and from 105 cos 15° = cos 45° cos 30°+sin 45° sin 30° 

Whence sin 15° = ^-^ = .258819 

2\/2 

and cos 15° = ^-i±i = .965926 

2V2 

Also, by 33 tan 1 5° = 2 - y/z = .267949 

and from 34 cot 15°=2+ \/3 = 3.732051 



60 PLANE TRIGONOMETRY 

123. Ratios for 18°. Let 18° =x. 

Then 3x = 54°, 2x = 36°, and 2x+3x = 90° 

Hence, by 38 cos3x = sin2x 

By 100 cos 3x = cos (2x+x) =cos 2x cos x— sin 2x sin x 
By 109 sin 2x = 2 sin x cos x and cos 2x = cos 2 x— sin 2 x 

Therefore cos 3x = (l— 4 sin 2 x) cos x 

Hence (1—4 sin 2 x) cos x = 2 sin x cos x 

Therefore 4 sin 2 x+2 sin x = 1 

From this quadratic equation there results 

sinx = |(— l^y/b) 
Whence sin 18° = j( V5-l) = .3 09017 

and by 32 cos 18° = J VlO+2 V5 = .951057 

-\A— l 
From 33 tan 18° = / = .324919 

V10+2V5 

From 34 cot 18° = V l0 +gVo = 3.077684 

V5-1 

124. Ratios for 3°. 

From 104 sin 3° = sin 18° cos 15°-cos 18° sin 15° 
and cos 3° = cos 18° cos 15°+sin 18° sin 15° 

Whence sin 3° = .052336 

and cos 3° = .998629 

Also, by 33 and 34 tan 3° = .052407 and cot 3° = 19.08113 

125. Ratios for 1° 30'. 



From 1 10 cos 1° 30' = yj 1+ ° 2 ° S 3 ° = .999657 

From 1 10 sin 1° 30" = Sm ,o OA „ = .026177 

2 cos 1 30 

By 33 and 34 tan 1° 30' = .026185 and cot 1° 30' = 38.18845 
126. Ratio for 45'. 



-i 



+cosl°30' =999914 



and from 1 10 sin 45' = ^ 1 °. 3 g ( f = .013090 

2 cos 45 

By 33 and 34 tan 45' = .013090 and cot 45' = 76.39000 



CALCULATION OF RATIOS 61 

127. Ratios for 1'. 

From 126, to the nearest sixth decimal, 

sin (45X1') = .013090 = tan (45X1') 
Therefore, to the nearest sixth decimal, 

tan (45 XI') -sin (45X1') is negligible. 

Whence by 118, 

s . nr = sin(45Xl') =000291 
45 

and by 1 19, tan l' = tan { f r XV) = .000291 

45 

Also by 32, cos 1' = 1.000000 

and by 34, cot 1' = 34.3477 

CALCULATION OF TABLES 

128. In formulas given in 1 13, 1 14, and 1 15 make 2=1'. 
Then 

sin n l' = 2 sin (n-l) V cos l'-sin (n-2) 1' 

cos n l'=2 cos (ft—1) 1' cos V— cos (n— 2) 1' 

, , sin n V , . , cos n V 

tan?ir = — and cotnr=- 



cos n V sin n V 

By giving successive values to n, ratios for angles at intervals 
of 1' may be calculated. And similarly for other intervals. This 
process, however, need not be carried beyond 30°, because by 108 

sin (30°+z) =2 sin 30° cos z-sin (30°-z) 
and by 121 2 sin 30° = 1 

Whence sin (30°+*) =cos z-sin (30°-z) 

Similarly cos (30°+z) =cos (30°-a;) -sin x 

Then, by 33 tan (30=+x).;^±| 

a„db y 34 cot^-I^M 

This last process need not be carried beyond 45°, because by 38 
sin (45°+x)=sin [90°-(45°-a;)]=cos (45°-x) 
cos (45°+x)=cos.[90°-(45°-*)]=sin (45°-x) 
tan (45°+x)=tan [90°-(45°-x)] = cot (45°-x) 

and cot (45°+x) = cot [90°-(45°-x)]=tan (45° -x) 

The process just explained is known as Simpson's Method of 

Calculating Trigonometrical Tables. 



CHAPTER XIII 



RATIOS OF ANGULAR MEASURE FOR ANY ANGLE 

129. To apply ratios of angular measure to angles of e very- 
possible magnitude requires that conventions of algebraic signs, as so 
far explained, shall be extended to all measurements of angles and 
lines made in opposite directions and considered with respect to 
each other. 

130. Let VAX be a plane angle and suppose a straight line 
to be rotated in plane VAX from position AZ to position A Y. 
Required the angle V A F. 

Let 

VAY=V, VAX=X, and YAX=Y 

Let all angles be regarded as positive and let the signs + and — 
denote only addition and 
subtraction. 

Suppose the line had 
been rotated from AZ to- 
ward AV, as in the upper 
diagram. 
Then V=X-Y (1) 

But suppose it had 
been rotated from AZ away 
from AV, as in the lower 
diagram. 
Then V=X+Y (2) 

Hence, to express the value of V requires a different formula 
for each case. 

But by Algebra 

(1) may be written V=X-(+Y) 

(2) may be written F=X-(-F) 

Now let Y be regarded as positive when in the direction of A V 
and as negative when in the opposite direction. 

Then in both cases V = X — Y 




RATIOS OF ANGULAR MEASURE FOR ANY ANGLE 63 



131. By virtue of the principles illustrated in 58, 59, 60, and 
130, and in order to make formulas for distance and direction gener- 
ally true, mathematicians have 
adopted the following general 
conventions: 



132. All measurements of 
angles and lines when consid- 
ered without regard to measure- 
ments in the opposite direction 
must be regarded as positive. 

133. All angles measured 
in the direction contrary to the 
motion of the hands of a clock 
must be regarded as positive. 

134. All angles measured in 
the same direction as the motion 
of the hands of a clock must be 
regarded as negative with respect 
to angles measured in the oppo- 
site direction. 

T35. All distances meas- 
ured to the right or upward must 
be regarded as positive. 

136. All distances meas- 
ured to the left or downward 
must be regarded as negative 
with respect to the distances 
measured in the opposite direc- 
tion. 




,,- » ■ 



These conventions are of great importance in all the higher 
branches of mathematics. 



64 



PLANE TRIGONOMETRY 




137. Ratios of angular measure for any angles are derived from 
the right triangle formed by drawing from a point on one side of the 
angle a perpendicular to 
the other side or to the 
other side prolonged. 

Let X'AX be a 
straight line of known 
position in any real or 
imaginary plane. With 
A as a center and AC as 
a radius describe a circle. 
Through A draw two 
diameters BE and DF, 
each making the same 
acute angle A with the initial line X'AX, and complete the four 
equal right triangles as shown. 
Then, measuring from AX over to the left 

XAB = A, XAD = 180°-A, XAE=180°+A, and XAF = 360°-A 

Let a, b, and h, respectively, denote the number of units of linear 
measure in the sides of each right triangle as shown. 

By 132 the radius, being taken regardless of measurement in 
the opposite direction, is positive for any position. 

By 133 the angles measured over to the left are positive. 

By 135 the distances AH, HB, and KD, being measured to 
the right and upward, are positive. 

By 136 the distance AK, HF, and KE, being measured to the 
left and downward, are negative. 

Therefore, by definition and convention 



sin A = -r 
n 



cos A =-r 
n 



tan A=-r- 
b 



cot A » 



sin (180°-A)=-|- cos (180°- A) =-^ tan (180° -A) =-^- cot (180°- A) =-^ 
sin (180°+A) =^ cos (180°+A) =^ tan (180°+A) =5| cot (180°+A) =^ 



sin (360°-A)=-^ cos (360°-A)=y 



tan(360°-A)=-~ cot (360° -A) =3^ 



RATIOS OF ANGULAR MEASURE FOR ANY ANGLE 65 



Whence sin A= sin (180°-A) = -sin (180°+A) = -sin (360°- A) 
cos A= -cos (180°- A) = -cos (180° -f- A) = cos (360°- A) 
tan A= -tan (180°- A) = tan (180°+A) = -tan (360°- A) 

and cot A= -cot (180°- A) = cot (180°+A) = -cot (360°- A) 

138. Since the formulas last given are true for any acute angle 
A, they are also true for the acute angle 90° — A. 

Whence sin A= -cos (90°+A) = -cos (270°- A) = cos (270°+A) 
cos A = sin (90°+A) = -sin (270°- A) = -sin (270°+A) 
tan A= -cot (90°+A) = cot (270°- A) = -cot (270°+A) 

and cot A- -tan (90°+A) = tan (270°- A) = -tan (270°+A) 

RATIOS FOR NEGATIVE ANGLES 

139. From any initial line X'A X, lay off over to the left the 
angle XA V and under to the left the angle XAZ, each angle con- 
taining A units of angular 
measure. 
Then, by 133 

XAV=+A 
and by 134 

XAZ=-A 
Layoff AB = AB' and 
draw BB', thus forming the 
two right triangles AHB 
and AHB', the sides of 
which determine the ratios for the angles A and —A. 
Then by 132, AB and AB f are positive; by 135, HB is positive; 
and by 136 A H and HB' are negative. 
Whence, for any two angles A and — A 

sin (— A) = — sin A cos (— A)= cos A 
tan(— A) = — tan A cot ( — A) = — cot A 

RATIOS FOR ANGLES GREATER THAN 360° 

140 If any angle is changed by adding or subtracting 360°, it 
is evident that the position of the right triangle from which the ratios 
are determined will remain unchanged, and the same is true if any 
number of times 360° is added or subtracted. 

Whence, if n is any integer and A any angle: 

Each ratio for ±A = same ratio for (±A^n 360°) 




66 PLANE TRIGONOMETRY 

EXAMPLES FOR PRACTICE 

Verify the following equations: 

1. sin 167° 29'= .2167 5. sin 223° 42'= -.6909 

2. cos 150° 38'= -.8715 6. cos 297° 34'= .4628 

3. tan 176° 19' = - .0644 7. tan 197° 56' = .3236 

4. cot 91° 45'= -.0306 8. cot 327° 10'= -1.5497 

GENERALIZATION OF FORMULAS 

141. Let A be an acute angle. Then 

From 138 sin (90°+ A) = cos A, cos (90°+^) = -sin A 

From 137 sin (180°+^) = -sin A, cos (180°+^) = -cos A 

From 138 sin (270°+^) = -cos A, cos (270°+^4) = sin A 

From 140 sin (360°+^)= sin A, cos (360°+^) = cos A 

142. Let A + B be an acute angle. 
Then, by 99, 100, and 141 

sin (90°+ A+B) = cos (A + B) = cos A cos B -sin A sin B 
sin (180°+ A+B) = -sin (A+B) = -sin A cos B-cos A sin B 
sin (270°+^4 + B) = -cos (A + B) = -cos A cos B+sin A sin B 
sin (360°+^+B) = sin (A+B) = sin A cos 5+cos A sin B 

And by substitution 

sin (90°+ A+B) =sin (90°+ A) cos B+ cos (W+A) sin B 
sin (180° +4+5) = sin (180°+ A) cos B+cos (180°+^) sin B 
sin (270°+^+B)=sin (270°+^) cos 5+cos (270°+^) sin B 
sin (360°+ 4+5) = sin (360° -M) cos 5+cos (360°+4) sin B 

Therefore, by 140, if n is any integer 
sin (n90°+A+B) =sin (n90°+A) cos B+cos (n90°+A) sin B 

This formula, evidently, will also be true if n.90° is added to B 
instead of to A and by similar processes it may be shown to be 
true when n90° is added to both A and B. 

Now let 

n90°+A=x and n90°+B=y 

Then, for any value of n and, therefore, for any value of x or y 

sin (x+y) = sin x cos y+cos x sin y 
Similarly cos (x+y) = cos x cos y— sin x sin y 



RATIOS OF ANGULAR MEASURE FOR ANY ANGLE 67 

tan x+t&n y 



Also, as in 101 tan (x+y) = 
and as in 102 cot (x+y) = 



1 — tana: tan y 
cotzcot?/— 1 



cot x + cot y 

Similarly for any values of x and y 

sin (x— y) = sin x cos y — cos x sin y 
cos (x—y) = cos x cos y+sin # sin y 
tana;— tan y 



tan (x—y) = 
and cot (#— y) = 



l+tanztany 

cotacoty+l 
cot 2/— cot x 



The same may be shown to be true of all general formulas of 
trigonometry. 

PROBLEMS FOR ILLUSTRATION 

143. 1. Find the ratios for the angle 470° 20'. 
sin 470°20' = sin (360°+110° 20') = sin 110° 20' 

= sin (180°- 69° 40')= sin 69° 40'= .9377 
cos 470°20' = cos (360° +110° 20')= cos 110° 20' 

= cos (180°- 69° 40 r ) = -cos 69° 40'=- .3475 
tan 470°20' = tan (360°+ 110° 200= tan 110° 20' 

=tan (180°- 69° 400 = -tan 69° 40'= -2.6985 
cot 470°20 , = cot (360°+110°200= cot 110° 20' 

-cot (180°- 69° 400 = -cot 69° 40'=- .3706 

2. Find all angles less than 360° for the sine ±.2700. 

A T+.2700 = sin 15° 40' = sin 164° 20' 

nswer \ _ .2700 = sin 195° 40' = sin 344° 20' 

3. Find all angles less than 360° for the cosine ±.8465. 

A ( + .8465 = cos 32° 10' = cos 327° 50' 

WCr i -.8465 = cos 147° 50' = cos 212° 10' 

4. Find all angles less than 360° for tangent =1.5697. 

A ( + 1 .5697 = tan 57° 30' = tan 237° 30' 

I -1.5697=tan 122° 30' = tan 302° 30' 

5. Find angles less than 360° for cotangent ±.1228. 

. r +.1228 = cot 83° 00' = cot 263° 00' 

Answer j _ ^ = ^ g?0 Q() , = ^ mQ 0Q/ 



CHAPTER XIV 

GENERAL THEORY OF CIRCULAR MEASURE 

144. Practical conditions require that angles shall be meas- 
ured by means of degrees, minutes and seconds or other sub- 
divisions of the circumference of a circle. For theoretical pur- 
poses, however, it is often convenient to express the magnitude 
of an angle by the ratio of the length of the subtending arc to 
the length of the radius. 

This ratio is known as the circular measure of the angle. 

145. Let YAX be a plane angle. 

With A as a center and any distance AB as a radius describe 
the arc of a circle intersecting AY at B and AX at C. 

Let a = linear length of 
arc BC 
r = length of the 

radius AB 
C = the circular 
measure of 
angle YAX 



Then 



r 




146. The Geometrical 
principles upon which Cir- 
cular Measure is based are 
as follows: 

147. Let YAX be sl 

plane angle. 

With i as a center 
and any radii r and r f , 
describe arcs of circles intersecting the sides of angle YA X as shown. 

Then, by Geometry, 



Whence: For the same angle the circular measure is constant. 



CIRCULAR MEASURE 



69 



148. Let YAX and 

VA X be any two angles in 
the same plane with a com- 
mon vertex A. 

With i as a center 
and with any radius r, 
describe the arc BDC of a 
circle intersecting the sides 
of the angles as in the n 
figure. 

Then by Geometry, 

arc BC : arc DC : : 

arc BC arc DC 
or, 




angle YAX : angle VAX 
angle YAX : angle VAX 



r r 

Whence: The circular measure of an angle is proportional 
to its magnitude. 

149. It is proved in Geometry that the ratio of the length 
of the semicircumference of a circle to the length of its diameter 
is 3.141,592,653,590, to the nearest twelfth decimal. This ratio 
is usually denoted by the Greek letter ir. 

Therefore, w is the circular measure of an angle represented 
by a semicircumference or 180°. 

150. Let A = degrees, minutes, and seconds in an angle and 
let C = the circular measure of that angle. 

Then by 148, 

C : ir :: A : 180 
or, 

C=-^-tt and A = — 180° 

180 7T 

PROBLEMS FOR ILLUSTRATION 

151. 1. Given A = 37° 47' 54", to find C. 

37° 47' 54" = 136,074" and 180° = 648,000" 

Therefore, C = ^|^tt = . 659706 

648,000 

2. Given (7 = 3.056765, to find A. 

A = l 141593 180 ° = 175 ° 8 ' 23 " 



70 



PLANE TRIGONOMETRY 



152. The angle of which the circular measure is unity is 
known as a radian. 

Let i2 = the degrees, minutes and seconds in a radian. 

Then, by ISO, R = — 180° = 57° 17' 45" 

w 

CIRCULAR MEASURE AND RATIOS OF ANGULAR MEASURE 

153. Let the acute angle 
YAX=A. 

With iasa center, describe 
the arc of a circle intersecting 
AY at B and AX at C. Draw 
B H and CD perpendicular to AX. 




— x 



Then, by Geometry BH<BC<CD 
Therefore 
BH 



BH < BC < CD 

r r " r 



BC CD 

Now ^^ = sin A, — = cir. meas. of A, and — =tan A 
rr r 

Hence sin ^4<cir. meas. of ^4<tan A 

and n sin A <n cir. meas. of A <n tan A 

But by 148 n cir. meas. A = cir. meas. nA 

Therefore n sin A <cir. meas. nA <n tan A 



RELATIONS FOR SMALL ANGLES 

154. If an angle nx is so small that tan nx — sin nx is 
negligible, [then by 117, 118, 119, and 153, for practical pur- 
poses, 
sin nx = n sin # = cir. meas. nx = n cir. meas. x = n tan # = tan nx 



or 



and 



sin nx 



cir. meas. nx 
sin x 






tan nx 



1 



cir. meas. nx 
tan x 



cir. meas. x cir. meas. x 

155. For illustration, let no: = 45'. 
Then, from 126, 

sin 45' = .013090 tan 45 r = .013090 



and by 150, cir. meas. 45' = : 



45 



- = .013090 



180X60 
Whence, sin 45' = cir. meas. 45' = tan 45', practically 



CIRCULAR MEASURE 



71 



RELATIONS FOR SMALL INCREASES IN ANQLES 
156. Let all angles be expressed in circular measure, let the 
angle x be increased by angle z as in figure, and let s be the 
increase in sin x due to increase of z in angle. 

Then, 5 = sin (x+z)— sin x 

and by 111, sin (x+z) — sin x = 2 cos (x+^z) sin \z 
or by 100, s = (cos x cos \z — sin x sin \z) 2 sin \z 

Now, fa 

FK-BH FG 



6' = 




r 




r 




sin 


2 = 


FL 
r 






2 = 


arc 


FB 






and 



By Geometry, 
FG<FL<arcFB 

Hence, s<sinz<z A~ r ~ c 

If z is less than any assignable quantity then, practically, 
by 118, 2 sin Js = sin 2; by 154 sin 2 = 2; and by 31 cos §2 = 1 
and sin x sin §2 is negligible. 

Therefore, dividing the value of s by 2 sin §2 = 2 and dimin- 
ishing 2 indefinitely, there results, 

s sin (x+z) — sin x ^ /z rr 

— = = cos a;=vl-sin 2 a; 



157. In the last equation let s= f -5 — ^ )• 



Then, - — 

z 



T- A+ *> 



=cos (t"^) 



and by 38, — = ' = sin A = V 1 — cos 2 ^l 

2 2 

Now, cos A— cos (A—z) is the decrease in cos (A—z) due to 
an increase of 2 in that angle. 

Whence, if u is the decrease in cosine corresponding to an 
infinitesimal increase of 2 in an angle x, 
u _cos (x+z)— cos x 
2 z 



- = —sin x= — \/l— cos 2 z 



72 PLANE TRIGONOMETRY 

158. Let angle x be increased by z and let t be the coi 
responding increase in tan x. 

Then, f=tan (x+z)— tan x 

j u <j? x / , \ 4. sin (x+z) sin z 

and by 33, tan(z-fz)— tan z = ) — — (■ 

cos (£+2) cos a: 

Reducing to a common denominator and applying 104 and IOC 

i=tan (x+z) — tan z = - : : r 

(cos x cos 2 — sin x sin 2) cos a: 

Whence by 156, if z is diminished indefinitely and if the valu 

of t is divided by sin 2=2 there results, 

t tan (x+z)— tan x 1 ;., ; , 
— = - = — — = l+tan 2 a; 

Z Z COS 2 £ 

7T 

159. In the last equation let x = — — A 
t ten(f~^+»)-tan(^-^ 



Then, — = 

2 



, , ^ t cot (A— z)— cot A 1 

and by 38, — = 



'M 



z 2 sin 2 J. 

Now, cot ^4 — cot (.4—2) is the decrease in cot 04—2) due t( 

an increase of 2 in that angle. So that, if v is the decrease ii 

cotangent corresponding to an infinitesimal increase of 2 in angle x 

v cot (x+z) — cot x 1 . x0 

— = i 1 = — __ -= — 1— cot 2 # 

2 2 sin 2 # 

160. The relations given in 156, 157, 158, and 159 are o 
great importance in the higher branches of mathematics. 

CALCULATION OF THE RATIOS FROM CIRCULAR MEASURE 

161. To express the ratios of angular measure of an angle ir 
terms of circular measure. Let x be the circular measure anc 
assume that for all values of x, 

sin x = a— bx+cx 2 +dx s +ex i +fx?+gx G +hx 7 + . . . 
By 139, —sin x = a — bx+cx 2 — dx z +ex* — fz*+gx* — hx 7 + . . . 
Subtracting and dividing by 2x, 

^^ = b+dx 2 +fx A +hx*+ . . . 

X 

sin x 
By 1 54, when x = 0, then = 1 . Consequently, 6 = 1 



CIRCULAR MEASURE 73 

Hence, sin x = x+dx 3 +fx b +hx 7 + 

and by hypothesis, 

sin (x+z) = (x+z)+d(x+z)*+f(x+z) 5 +h(x+z) 7 + 

Expanding (x+z), subtracting sin x and reducing, 

sin (x+z) — sin x = 

z+d(3x 2 z+3xz 2 +)+f(5x*z+10xh 2 +)+h(7x 6 z+21x 5 z 2 +) + 

Dividing out by z and then making it infinitesimal, 

sin 0+z)-sin x , , 4 _ 7 6 

= l+3az 2 +of:r+//zx , + 

By 156, when z is less than any assignable quantity, 

sin (x +z)— sin x 
= cos X 

z 

Therefore, cos x = l+3dx 2 +5fx*+7hx 6 + .... 
and cos(x+z) = l+3d(x+z) 2 +5f(x+z) A +7h(x+z) G + . . . 
Whence cos (x+z) — cos x = 

3d(2xz+z 2 )+5f(4xh+6x 2 z 2 +)+7h (6x 5 z+l5x*z 2 +) + 
Dividing out by z and then making it infinitesimal, 

cos (x+z)— COS X . 7 , __- , , AC%1 z\ 

= 6dx+20fx*+42hx b + .... 

By 157, when z is less than any assignable quantity, 

cos (x+z) — COS X 

— — = -sin x 

z 

Hence, sin x= — Qdx — 20fx z — 42/b 5 — .... 

As before shown, sin x = x+dx z +fx h + 

Equating coefficients of like powers of x, there results 

6 J 120 5040 

Since 6 = 1X2X3, 120 = 1X2X3X4X5, 5040 = 1X2X3X4X5 
X6X7, they and similar products will be denoted by the factorial 
sign, | which is generally used for that purpose. 

/v»3 <j»5 /v»7 />»" 

Whence, sinx=x— ■ nr+r? — t^t+tt; 1" • • • 

Li IA l-L Li. 

X 2 X 4 X 6 X 8 

COSX = l-|Y+y^-^-+|y-+ . . . 

From 33 tanx = x+^-+— x 5 + — -x 7 +— -x 9 + . . . 
6 15 olo 2o6o 

X X^ 2 T 

and by 34 cotx= — — — rjrX 5 — 



x 3 45 945 4725 



74 



PLANE TRIGONOMETRY 



7T 



162. To calculate the ratios for — or 18° by the formulas 
given m 161. 

This involves the use of the following powers of 7r: 
tt = 3.141593 tt 2 = 9.869604 tt* = 31.006277 



tt 4 = 97.409091 7T 5 = 306.019685 


7T 6 = 


= 961.389194 


tt 7 = 3020.293228 


i 

7T 


.3183098 




or o*^ o* 
1. Sh*-*-jy+|y-jy- + 

Here, x = 




- • • 




.314159 


x z 








-.005168 


x b 

\T 

~J7 








.000026 


X 1 








-.00000006 



Whence, as in 123, sin 18° = 

/y«2 />*4 /y»6 

2. Cos x=l 



Here, 1 = 



ll T I±. \± 









16 



Whence, as in 123, cos 18° = 



62 



3. Tto*-*+f +A x . + ^ /+ _ i _ : 



Z 9 + 



.309017 

1.000000 
-.049348 

.000406 

-.000001 

.951057 



Whence, by process just indicated, there results, 
as in 123, tan 18°= .324919 



1 



x 



X? 



X' 



4 ' Cot X x 3 45 945 a 4725 
Whence, there results, as in 123, cot 18° = 



3.077684 



CHAPTER XV 

INVERSE RATIOS OF ANGULAR MEASURE 

163. The symbols, sin" 1 , cos -1 , tan"" 1 , cot -1 , etc., are used to 
express an angle in terms of its ratios of angular measure, it being 
understood that as so used ~Y is not a negative exponent. 

Thus, the equation x = sin _1 2/ means that x is an angle whose 
sine is y or that y = sin x. 

In this connection it must be borne in mind that while there 
is but one set of ratios for any given angle there are many angles 
corresponding to any given ratio. Thus, from annexed table, 
by 67, 

sin-^OOO is 30°, or — and also 150°, or — -ir 
6 6 

and by 137, sin" 1 -. 5000 is 210°, or -Lr and also 330°, or -^tt 

6 6 

Similarly for the other ratios. 

164. Let x be an angle whose sine is y, or y = sin x. 

Then by 38, 2/ = cos (-^ — x 

Therefore, siir l y = x and cos~ l y = — — x 
and cos -1 ?/ = — — sin -1 ?/ 

Li 

Similarly, cot -1 ?/ = — — tan -1 !/ 

165. From 139, — sin -1 ?/ =sin _1 (— y) 
and — tan -1 ?/ = tan -1 (—y) 

166. Let tan x = m and tan z = n. 
Then, z = tan -1 m and 2 = tan -1 ?t 

m+n 



By 101, tan(z+z) = 



1—mn 



Whence, £+z = tan -1 

1 — mn 

and tan _1 m+ tan _1 n = tan -1 :; 

1—mn 

Also by 106, tan -1 m — tan -1 ™ = tan" 1 — ^— 

1+mn 



76 PLANE TRIGONOMETRY 

RELATIONS FOR SMALL INCREASES IN RATIOS 

167. Let sin x be increased by s and let z be the correspond- 
ing increase in the angle x. 

Let y = sin x. Consequently, ?/+s = sin (x+z) 
Then, x = siir 1 y and x+z = sin~ 1 (y+s) 

Also, s = sin (x+z) — sin x z = sin~ 1 (y+s) — sin -1 # 

s _sin (x+s) — sinx _z__sirr 1 (y+s) — sm~h/ 
z z s s 

By Geometry, when s is less than any assignable quantity, z is 
infinitesimal and by 156 when z is infinitesimal, 

s sin (x+s) — sin x /z r— 

— = = cos a;=vl-sin 2 a; 

z z 

Therefore, when s is less than any assignable quantity, 

z _sm~ l (y+s)—sm~ 1 y _ 1 _ 1 

s s y/\ — sin 2 x -\/l—y 2 

Whence, by the Binomial Theorem, 

sm-\y+s)-sm-iy _ y> 1X3 4 1X3X5 6 1X3X5X7 8 

.9 2 2xr 2X4X6^ 2X4X6X8^ ' ' 

168. Let tan x be increased by t and let z be the correspond- 
ing increase in the angle x. 

Let ?/ = tan x. Consequently, ?/+£ = tan (x+z) 
Then, x = tan -1 ?/ and x+z = tsLir 1 (y+t) 

Also, 2 = tan (x+s)— tan x z = tsar 1 (y+t)—t8irr 1 y 

t _tan (x+z)— tan x z _tan -1 (?/+f) — tan -1 ?/ 

By Geometry, when t is less than any assignable quantity, z is 
infinitesimal and by 158, when z is infinitesimal, 

t tan (x+z)— tan x t 2 

— == — — = 1 +tan 2 x = 1 +y 2 

z z 

Therefore, when t is less than any assignable quantity, 

z _ tan -1 (?/+f) — tan -1 ?/ _ 1 _ 1 
T~ ~T~ ~l+tan 2 x~l+?/ 2 

Whence, by division or by the Binomial Theorem, 

tan -1 (?/+£) — tan -1 ?/ _, , , * fi . . 
^Y = 1— 2/ 2 H-2/ 4 — 2/ 6 H-2/ s — • . • 



INVERSE RATIOS OF ANGULAR MEASURE 77 

CALCULATION OF CIRCULAR MEASURE FROM THE RATIOS 

169. To express the circular measure of sirr l y and of cos _1 y 
in terms of the sine and cosine. 

Assume that for all values of y, 
sm-iy = A + By+Cy*+Dtf+Ey*+Fy 5 +Gy*+Hy 7 +Iy*+Jy 9 + . . 

From 165, — sin~ 1 i/ = sin~ 1 (— y) 

Hence, by substitution, 
-sm-iy = A-By+Cy*-Dy*+Ey*-Fy*+Gy«-Hy 7 +Iy*-Jy>+ 

Subtracting and dividing out by 2y there results, 

*™-l = B+Dtf+Fy*+Htf+Jy*+ . . . 

y 

sin — ^t/ 
By 31 and 154, when sin -1 r/ = 0, then y = and -=1. 

y 

Therefore, 5 = 1 

and sm- l y=y+Dy s +Fy i +Hy , '+Jy°+ .... 

Since this equation is true for all values of y, 
sin- 1 (y+s)=y+s+Df +Fy*+ 

or, 
sin-^+s) = y+s+Dy 3 +D(3y*s+3ys*+) +Fy i +F(5y*s+ 10t/¥+) + 

Subtracting the last value of sin -1 ?/, dividing out by s, and 
then making s infinitesimal, 

Sm ~ 1(y+s) ~ sin ~ ly = l+3Dy*+5Fy*+7Htf+9Jf+ . . . 
s 

By 167, when s is infinitesimal, 
sin-Hy+^Q-sin-iy y* 1 X 3 . 1X3X5 . 1X3X5X7 . 
s 2 2X4* 2X4X6* 2X4X6X8* 

Equating coefiicients of like powers of y in the last two equa- 
tions and reducing, there results, 

£=— F=— H=— J=- Zb 



6 40 112 1152 

Whence, &nr t y=y+lL+ ^+J^+^f+ ■ ■ ■ 

By 164, cos" 1 2/ = "^ — sin _1 t/ 

Whence , cos -1 v = v — v 5 v 7 v 9 — 

i y 2 j 6 4Q 2/ 112 ^ llg2 y 



78 PLANE TRIGONOMETRY 

170. To express the circular measure of tan -1 ?/ and of cot -1 y 
in terms of the tangent and cotangent. 

Assume tan~ 1 y = A+ By+Cy 2 +Dy s +Ey i +Fy 6 +Gy e +Hy''+. 
Then, from 165, 31, 154, by processes shown in 169 with 
respect to the sine and by 168, 

tan-»y = y+ W+ Ff+ Hy 7 + . . . 
tan-t(y+Q-taii-y Ml+ 3 IV+5Fyt+7gpt+ _ _ _ 

and tan-*W)-tan-V = 1 _ f+ yi _ y+ . . . 

Equating coefficients of like powers of y in the last two equations, 

p __i, ,_£ H ._| 

Whence, tan"ty = y — ^-y s +Ty* — =V+ . . 

6 5 7 

and from 164, cot- l y = —T-y+—y z — —y b +-=-y 7 - . . 

Z 6 7 

PROBLEMS FOR ILLUSTRATION 

171. To calculate the circular measures of sin~~ 1 .052336, cos -1 
.052336, tan~ 1 .052407 and cot" 1 .052407. 

1 3 

1. From 169, sm~ 1 y=y+—y*+—y*+ 

Whence, sin" 1 .052336 = .052360 

2. By 164, cos- 1 .052336 = -^7r-sin- 1 .052336= 1.518436 

3. From 170, tan~ 1 y=y — —y 3 +—y b - . • 

o 5 

Whence, tan" 1 .052407 = .052360 

4. 'By 164, cot- 1 .052407 =-Ur-tan- 1 .052407 = 1.518436 

Zi 

5. To verify the results thus obtained, 

By 124, sin- 1 .052336 = tan- 1 .052407= 3° 

From 38, cos- 1 .052336 = cot- 1 .052407 = 87° 
By 149, cir. meas. 3° = .052360 cir. meas. 87°= 1.518436 



CHAPTER XVI 

SUPPLEMENT 

172, In addition to the sine, cosine, tangent, and cotangent, 
there are certain other ratios of the same kind which, for com- 
pleteness, will now be explained in general terms. 

SECANT AND COSECANT 

173. In any right plane triangle ABH with sides and angles 

as shown, 

£ A hypotenuse A h 

secant of A = . ., , . or sec A = — 

side adjacent b 

« A hypotenuse A h 

cosecant of A = . , : — or cosec A= — 

side opposite 

By 28, sec A = 



cosec A — 



cos A 
1 




sin A 

174. From these definitions and relations and from 31, the 
limiting values of the secant and cosecant are, 

sec 0° = -x = unity, sec 90° = —£ = infinity 

cos cos 90 

cosec 0° = - — -h = infinity, cosec 90°= r^= unity 

sin , cos 90 

175. Since the secant is the reciprocal of the cosine their 
algebraic signs are the same for the same angle. Similarly, for the 
same angle the algebraic signs of the cosecant and sine are the same. 

EXAMPLES FOR PRACTICE 

176. Prove the following relations. 

1. sec ^4 = cosec(90° — ^4). 2. cosec ^t = sec(90° — A). 

3. sec A sin A = tan A. 4. cosec A cos A = cot A. 

5. sec 2 ^4 = l+tan 2 ^4. 6. cosec 2 ^4 — cot 2 ^4 = l. 

7. sec A — tan A= 7 

tan A+ sec A 

8. sec ^l = cosec (90°+^) = -sec (180°+^)«sec(360°-^). 

9. sec ( — A)*=sec A, 10. cosec (—A) «= —cosec A. 



80 



PLANE TRIGONOMETRY 



VERSED SINE AND COVERSED SINE 



177. Through a point A 
draw an initial line XX' and 
lines VV f and ZZ' making same 
acute angle A with XX'. 

With A as a center and r 
as a radius describe a circle and 
complete figure as shown. 

Then, 

HC 



and 



versed sine 


of A 


= - 


r 


coversed sine of A 




NL 

r 


vers 


( 90°- 


■A) 


_NL 
r 


vers 


(180°- 


■A) 


_KC 
r 


vers 


(180°+^) 


_KC 

r 


vers 


(360°- 


■A) 


_HC 
r 




covers ( 90° — A) = 
covers (180° -4) = 
covers (180°+^) = 



HC 

r 
NL 

r 
OL 

r 

covers (360°-^)= — 
r 

Now, HC = AC-AH, NL = AC-HB, OM = NL 

KC = AC+AK = AC-(-AH), 0L = AC-(-HB). 

Whence by 28, 66, 137, 138, 139 and 140, for any angle, 

versed sine of A =1 — cos A 

coversed sine of A = l — sin A. 

178. The limiting values of the versed sine and coversed sine 
are as follows, 

vers 0° = l-cos 0° = 0, vers 180° = 1 -cos 180° = 2 
covers 90° = 1 -sin 90° = 0, covers 270° = 1 -sin 270° = 2 

179. Since the value of the sine or of the cosine can never 
exceed unity the algebraic signs of all versed sines and coversed 
sines must be positive. 

EXAMPLES FOR PRACTICE 

180. 1. Verify, vers A sec A = sec A — l 

2. Prove that, covers A(2 — covers A)= cos 2 A 

3. Show that, vers A + (1— covers ^4)cot A — l. 



TABLE OF 

NATURAL SINES, COSINES 

TANGENTS AND COTANGENTS 



TABLE OF 

NATURAL SINES, COSINES, TANGENTS, AND COTANGENTS 
(Advanced by 10 Minutes) 











+* 










■*» 


Degrees 

Minutes 


m 


a 

•5! 

o 
O 


a 

0) 

a 
3 
H 


a 

03 

■+* 

o 
O 


* & 
Q S 


1 


•S 
"3 

6 


<y 


c 

a 

3 

-♦* 

o 
O 


00 00 

10 
20 


.0000 
.0029 
.0058 


one 
one 
one 


.0000 
.0029 
.0058 


Infinite 
343.77 
171.89 


6 00 
10 
20 


.1045 
.1074 
.1103 


.9945 
.9942 
.9939 


.1051 
.1080 
.1110 


9.5144 
9.2553 
9.0098 


30 
40 
50 


.0087 
.0116 
.0145 


one 
.9999 
.9999 


.0087 
.0116 
.0145 


114.59 
85.940 
68.750 


30 
40 
50 


.1132 
.1161 
.1190 


.9936 
.9932 
.9929 


.1139 
.1169 
.1198 


8.7769 
8.5555 
8.3450 


1 00 
10 
20 


.0175 
.0204 
.0233 


.9999 
.9998 
.9997 


.0175 
.0204 
.0233 


57.290 
49.104 
42.964 


7 00 
10 
20 


.1219 
.1248 
.1276 


.9925 
.9922 
.9918 


.1228 
.1257 
.1287 


8.1443 
7.9530 
7.7704 


30 
40 
50 


.0262 
.0291 
.0320 


.9997 
.9996 
.9995 


.0262 
.0291 
.0320 


38.188 
34.368 
31.242 


30 
40 
50 


.1305 
.1334 
.1363 


.9914 
.9911 
.9907 


.1317 
.1346 
.1376 


7.5958 

7.4287 
7.2687 


2 00 
10 
20 


.0349 
.0378 
.0407 


.9994 
.9993 
.9992 


.0349 
.0378 
.0407 


28.636 
26.432 
24.542 


8 00 
10 
20 


.1392 
.1421 
.1449 


.9903 
.9899 
.9894 


.1405 
.1435 
.1465 


7.1154 
6.9682 
6.8269 


30 
40 
50 


.0436 
.0465 
.0494 


.9990 
.9989 
.9988 


.0437 
.0466 
.0495 


22.904 
21.470 
20.206 


30 
40 
50 


.1478 
.1507 
.1536 


.9890 
.9886 
.9881 


.1495 
.1524 
.1554 


6.6912 
6.5606 
6.4348 


3 00 

10 
20 


.0523 
.0552 
.0581 


.9986 
.9985 
.9983 


.0524 
.0553 
.0582 


19.081 
18.075 
17.169 


9 00 

10 
20 


.1564 
.1593 
.1622 


.9877 
.9872 
.9868 


.1584 
.1614 
.1644 


6.3138 
6.1970 
6.0844 


30 
40 
50 


.0610 
.0640 
.0669 


.9981 
.9980 
.9978 


.0612 
.0641 
.0670 


16.350 
15.605 
14.924 


30 
40 
50 


.1650 
.1679 
.1708 


.9863 
.9858 
.9853 


.1673 
.1703 
.1733 


5.9758 
5.8708 
5.7694 


4 00 
10 
20 


.0698 
.0727 
.0756 


.9976 
.9974 
.9971 


.0699 
.0729 
.0758 


14.301 
13.727 
13.197 


10 00 
10 
20 


.1736 
.1765 
.1794 


.9848 
.9843 
.9838 


.1763 
.1793 
.1823 


5.6713 
5.5764 
5.4845 


30 
40 
50 


.0785 
.0814 
.0843 


.9969 
.9967 
.9964 


.0787 

.0816 
.0846 


12.706 
12.251 
11.826 


30 
40 
50 


.1822 
.1851 
.1880 


.9833 
.9827 
.9822 


.1853 
.1883 
.1914 


5.3955 
5.3093 
5.2257 


5 00 
10 
20 


.0872 
.0901 
.0930 


.9962 
.9959 
.9957 


.0875 
.0904 
.0934 


11.430 
11.059 
10.712 


11 00 

10 
20 


.1908 
.1937 
.1965 


.9816 
.9811 
.9805 


.1944 
.1974 
.2004 


5.1446 
5.0658 
4.9894 


30 
40 
50 


.0958 
.0987 
.1016 


.9954 
.9951 
.9948 


.0963 
.0992 
.1022 


10.385 
10.078 

9.7882 


30 
40 
50 


.1994 
.2022 
.2051 


.9799 
.9793 
.9787 


.2035 
.2065 
.2095 


4.9152 
4.8430 
4.7729 



TABLES 



83 



NATURAL SINES, COSINES, TANGENTS, AND COTANQENTS 
(Advanced by 10 Minutes) 



Degrees 
Minutes 


1 


| 

8 


1 
I 

§ 


a 

9 

M 

a 

o3 
O 

O 


Degrees 
Minutes 


1 
da 


1 

8 

o 


-** 
a 
5 

m 

a 

03 

h 


■4J 

a 

a> 

M 

a 

03 
O 
O 


12 00 


.2079 


.9781 


.2126 


4.7046 


18 00 


.3090 


.9511 


.3249 


3.0777 


10 


.2108 


.9775 


.2156 


4.6382 


10 


.3118 


.9502 


.3281 


3.0475 


20 


.2136 


.9769 


.2186 


4.5736 


20 


.3145 


.9492 


.3314 


3.0178 


30 


.2164 


.9763 


.2217 


4.5107 


30 


.3173 


.9483 


.3346 


2.9887 


40 


.2193 


.9757 


.2247 


4.4494 


40 


.3201 


.9474 


.3378 


2.9600 


50 


.2221 


.9750 


.2278 


4.3897 


50 


.3228 


.9465 


.3411 


2.9319 


13 00 


.2250 


.9744 


.2309 


4.3315 


19 00 


.3256 


.9455 


.3443 


2.9042 


10 


.2278 


.9737 


.2339 


4.2747 


10 


.3283 


.9446 


.3476 


2.8770 


20 


.2306 


.9730 


.2370 


4.2193 


20 


.3311 


.9436 


.3508 


2.8502 


30 


.2334 


.9724 


.2401 


4.1653 


30 


.3338 


.9426 


.3541 


2.8239 


40 


.2363 


.9717 


.2432 


4.1126 


40 


.3365 


.9417 


.3574 


2.7980 


50 


.2391 


.9710 


.2462 


4.0611 


50 


.3393 


.9407 


.3607 


2.7725 


14 00 


.2419 


.9703 


.2493 


4.0108 


20 00 


.3420 


.9397 


.3640 


2.7475 


10 


.2447 


.9696 


.2524 


3.9617 


10 


.3448 


.9387 


.3673 


2.7228 


20 


.2476 


.9689 


.2555 


3.9136 


20 


.3475 


.9377 


.3706 


2.6985 


30 


.2504 


.9681 


.2586 


3.8667 


30 


.3502 


.9367 


.3739 


2.6746 


40 


.2532 


.9674 


.2617 


3.8208 


40 


.3529 


.9356 


.3772 


2.6511 


50 


.2560 


.9667 


.2648 


3.7760 


50 


.3557 


.9346 


.3805 


2.6279 


15 00 


.2588 


.9659 


.2679 


3.7321 


21 00 


.3584 


.9336 


.3839 


2.6051 


10 


.2616 


.9652 


.2711 


3.6891 


10 


.3611 


.9325 


.3872 


2.5826 


20 


.2644 


.9644 


.2742 


3.6470 


20 


.3638 


.9315 


.3906 


2.5605 


30 


.2672 


.9636 


.2773 


3.6059 


30 


.3665 


.9304 


.3939 


2.5386 


40 


.2700 


.9628 


.2805 


3.5656 


40 


.3692 


.9293 


.3973 


2.5172 


50 


.2728 


.9621 


.2836 


3.5261 


50 


.3719 


.9283 


.4006 


2.4960 


16 00 


.2756 


.9613 


.2867 


3.4874 


22 00 


.3746 


.9272 


.4040 


2.4751 


10 


.2784 


.9605 


.2899 


3.4495 


10 


.3773 


.9261 


.4074 


2.4545 


20 


.2812 


.9596 


.2931 


3.4124 


20 


.3800 


.9250 


.4108 


2.4342 


30 


.2840 


.9588 


.2962 


3.3759 


30 


.3827 


.9239 


.4142 


2.4142 


40 


.2868 


.9580 


.2994 


3.3402 


40 


.3854 


.9228 


.4176 


2.3945 


50 


.2896 


.9572 


.3026 


3.3052 


50 


.3881 


.9216 


.4210 


2.3750 


17 00 


.2924 


.9563 


.3057 


3.2709 


23 00 


.3907 


.9205 


.4245 


2.3559 


10 


.2952 


.9555 


.3089 


3.2371 


10 


.3934 


.9194 


.4279 


2.3369 


20 


.2979 


.9546 


.3121 


3.2041 


20 


.3961 


.9182 


.4314 


2.3183 


30 


.3007 


.9537 


.3153 


3.1716 


30 


.3987 


.9171 


.4348 


2.2998 


40 


.3035 


.9528 


.3185 


3.1397 


40 


.4014 


.9159 


.4383 


2.2817 


50 


.3062 


.9520 


.3217 


3.1084 


50 


.4041 


.9147 


.4417 


2.2637 



84 



PLANE TRIGONOMETRY 



NATURAL SINES, COSINES, TANGENTS, AND COTANQENTS 
(Advanced by 10 Minutes) 



Degrees 
Minutes 


0> 

a 


.9 

O 

O 


1 

M 

a 

09 


•*> 

bO 

a 

03 

o 
O 


Degrees 
Minutes 




.5 
o 
O 


* 

1 


+3 

G , 
o3 

O 

U 


24 00 


.4067 


.9135 


.4452 


2.2460 


30 00 


.5000 


.8660 


.5774 


1.7321 


10 


.4094 


.9124 


.4487 


2.2286 


10 


.5025 


.8646 


.5812 


1.7205 


20 


.4120 


.9112 


.4522 


2.2113 


20 


.5050 


.8631 


.5851 


1.7090 


30 


.4147 


.9100 


.4557 


2.1943 


30 


.5075 


.8616 


.5890 


1.6977 


40 


.4173 


.9088 


.4592 


2.1775 


40 


.5100 


.8601 


.5930 


1.6864 


50 


.4200 


.9075 


.4628 


2.1609 


50 


.5125 


.8587 


.5969 


1.6753 


25 00 


.4226 


.9063 


.4663 


2.1445 


31 00 


.5150 


.8572 


.6009 


1.6643 


10 


.4253 


.9051 


.4699 


2.1283 


10 


.5175 


.8557 


.6048 


1.6534 


20 


.4279 


.9038 


.4734 


2.1123 


20 


.5200 


.8542 


.6088 


1.6426 


30 


.4305 


.9026 


.4770 


2.0965 


30 


.5225 


.8526 


.6128 


1.6319 


40 


.4331 


.9013 


.4806 


2.0809 


40 


.5250 


.8511 


.6168 


1.6212 


50 


.4358 


.9001 


.4841 


2.0655 


50 


.5275 


.8496 


.6208 


1.6107 


26 00 


.4384 


.8988 


.4877 


2.0503 


32 00 


.5299 


.8480 


.6249 


1.6003 


10 


.4410 


.8975 


.4913 


2.0353 


10 


.5324 


.8465 


.6289 


1.5900 


20 


.4436 


.8962 


.4950 


2.0204 


20 


.5348 


.8450 


.6330 


1.5798 


30 


.4462 


.8949 


.4986 


2.0057 


30 


.5373 


.8434 


.6371 


1.5697 


40 


.4488 


.8936 


.5022 


1.9912 


40 


.5398 


.8418 


.6412 


1.5597 


50 


.4514 


.8923 


.5059 


1.9768 


50 


.5422 


.8403 


.6453 


1.5497 


27 00 


.4540 


.8910 


.5095 


1.9626 


33 00 


.5446 


.8387 


.6494 


1.5399 


10 


.4566 


.8897 


.5132 


1.9486 


10 


.5471 


.8371 


.6536 


1.5301 


20 


.4592 


.8884 


.5169 


1.9347 


20 


.5495 


.8355 


.6577 


1.5204 


30 


.4617 


.8870 


.5206 


1.9210 


30 


.5519 


.8339 


.6619 


1.5108 


40 


.4643 


.8857 


.5243 


1.9074 


40 


.5544 


.8323 


.6661 


1.5013 


50 


.4669 


.8843 


.5280 


1.8940 


50 


.5568 


.8307 


.6703 


1.4919 


28 00 


.4695 


.8829 


.5317 


1.8807 


34 00 


.5592 


.8290 


.6745 


1.4826 


10 


.4720 


.8816 


.5354 


1.8676 


10 


.5616 


.8274 


.6787 


1.4733 


20 


.4746 


.8802 


.5392 


1.8546 


20 


.5640 


.8258 


.6830 


1.4641 


30 


.4772 


.8788 


.5430 


1.8418 


30 


.5664 


.8241 


.6873 


1.4550' 


' 40 


.4797 


.8774 


.5467 


1.8291 


40 


.5688 


.8225 


.6916 


1.4460 


50 


.4823 


.8760 


.5505 


1.8165 


50 


.5712 


.8208 


.6959 


1.4370 


29 00 


.4848 


.8746 


.5543 


1.8040 


35 00 


.5736 


.8192 


.7002 


1.4281 


10 


.4874 


.8732 


.5581 


1.7917 


10 


.5760 


.8175 


.7046 


1.4193 


20 


.4899 


.8718 


.5619 


1.7796 


20 


.5783 


.8158 


.7089 


1.4106 


30 


.4924 


.8704 


.5658 


1.7675 


30 


.5807 


.8141 


.7133 


1.4019 


40 


.4950 


.8689 


.5696 


1.7556 


40 


.5831 


.8124 


.7177 


1.3934 


50 


.4975 


.8675 


.5735 


1.7437 


50 


.5854 


.8107 


.7221 


1.3848 



TABLES 



85 



NATURAL SINES, COSINES, TANGENTS, AND COTANGENTS 
(Advanced by 10 Minutes) 



Degrees 
Minutes 


s 

a 
m 


© 
a 
"S3 
o 
O 


d 
5 

H 

a 

o3 

H 


9 

M 

q 

o3 
O 

O 


Degrees 
Minutes 


d 

w 


© 

.5 

"S3 

o 
O 


C 

a 

o3 


d 

0) 
CD 

C 
03 

O 

O 


36 00 


.5878 


.8090 


.7265 


1.3764 


42 00 


.6691 


.7431 


.9004 


1.1106 


10 


.5901 


.8073 


.7310 


1.3680 


10 


.6713 


.7412 


.9057 


1.1041 


20 


.5925 


.8056 


.7355 


1.3597 


20 


.6734 


.7392 


.9110 


1.0977 


30 


.5948 


.8039 


.7400 


1.3514 


30 


.6756 


.7373 


.9163 


1.0913 


40 


.5972 


.8021 


.7445 


1.3432 


40 


.6777 


.7353 


.9217 


1.0850 


50 


.5995 


.8004 


.7490 


1.3351 


50 


.6799 


.7333 


.9271 


1.0786 


37 00 


.6018 


.7986 


.7536 


1.3270 


43 00 


.6820 


.7314 


.9325 


1.0724 


10 


.6041 


.7969 


.7581 


1.3190 


10 


.6841 


.7294 


.9380 


1.0661 


20 


.6065 


.7951 


.7627 


1.3111 


20 


.6862 


.7274 


.9435 


1.0599 


30 


.6088 


.7934 


.7673 


1.3032 


30 


.6884 


.7254 


.9490 


1.0538 


40 


.6111 


.7916 


.7720 


1.2954 


40 


.6905 


.7234 


.9545 


1.0477 


50 


.6134 


.7898 


.7766 


1.2876 


50 


.6926 


.7214 


.9601 


1.0416 


38 00 


.6157 


.7880 


.7813 


1.2799 


44 00 


.6947 


.7193 


.9657 


1.0355 


10 


.6180 


.7862 


.7860 


1.2723 


10 


.6967 


.7173 


.9713 


1.0295 


20 


.6202 


.7844 


.7907 


1.2647 


20 


.6988 


.7153 


.9770 


1.0235 


30 


.6225 


.7826 


.7954 


1.2572 


30 


.7009 


.7133 


.9827 


1.0176 


40 


.6248 


.7808 


.8002 


1.2497 


40 


.7030 


.7112 


.9884 


1.0117 


50 


.6271 


.7790 


.8050 


1.2423 


50 


.7050 


.7092 


.9942 


1.0058 


39 00 


.6293 


.7771 


.8098 


1.2349 


45 00 


.7071 


.7071 


one 


one 


10 


.6316 


.7753 


.8146 


1.2276 


10 


.7092 


.7050 


1.0058 


.9942 


20 


.6338 


.7735 


.8195 


1.2203 


20 


.7112 


.7030 


1.0117 


.9884 


30 


.6361 


.7716 


.8243 


1.2131 


30 


.7133 


.7009 


1.0176 


.9827 


40 


.6383 


.7698 


.8292 


1.2059 


40 


.7153 


.6988 


1.0235 


.9770 


50 


.6406 


.7679 


.8342 


,1.1988 


50 


.7173 


.6967 


1.0295 


.9713 


40 00 


.6428 


.7660 


.8391 


1.1918 


46 00 


.7193 


.6947 


1.0355 


.9657 


10 


.6450 


.7642 


.8441 


1.1847 


10 


.7214 


.6926 


1.0416 


.9601 


20 


.6472 


.7623 


.8491 


1.1778 


20 


.7234 


.6905 


1.0477 


.9545 


30 


.6494 


.7604 


.8541 


1.1708 


30 


.7254 


.6884 


1.0538 


.9490 


40 


.6517 


.7585 


.8591 


1.1640 


40 


.7274 


.6862 


1.0599 


.9435 


50 


.6539 


.7566 


.8642 


1.1571 


50 


.7294 


.6841 


1.0661 


.9380 


41 00 


.6561 


.7547 


.8693 


1.1504 


47 00 


.7314 


.6820 


1.0724 


.9325 


10 


.6583 


.7528 


.8744 


1.1436 


10 


.7333 


.6799 


1.0786 


.9271 


20 


.6604 


.7509 


.8796 


1.1369 


20 


.7353 


.6777 


1.0850 


.9217 


30 


.6626 


.7490 


.8847 


1.1303 


30 


.7373 


.6756 


1.0913 


.9163 


40 


.6648 


.7470 


.8899 


1.1237 


40 


.7392 


.6734 


1.0977 


.9110 


50 


.6670 


.7451 


.8952 


1.1171 


50 


.7412 


.6713 


1.1041 


.9057 



86 



PLANE TRIGONOMETRY 



NATURAL SINES, COSINES, TANGENTS, AND COTANGENTS 
(Advanced by 10 Minutes) 



Degrees 
Minutes 


o 

QQ 


.9 

*SQ 

o 
O 


§ 

§ 


a 
o 

a 

08 
-♦* 

o 
O 


Degrees 
Minutes 


3 


a 

8 

O 


§ 

03 


c 

M 

a 

c3 

o 

O 


48 00 


.7431 


.6691 


1.1106 


.9004 


54 00 


.8090 


.5878 


1.3764 


.7265 


10 


.7451 


.6670 


1.1171 


.8952 


10 


.8107 


.5854 


1.3848 


.7221 


20 


.7470 


.6648 


1.1237 


.8899 


20 


.8124 


.5831 


1.3934 


.7177 


30 


.7490 


.6626 


1.1303 


.8847 


30 


.8141 


.5807 


1.4019 


.7133 


40 


.7509 


.6604 


1.1369 


.8796 


40 


.8158 


.5783 


1.4106 


.7089 


50 


.7528 


.6583 


1.1436 


.8744 


50 


.8175 


.5760 


1.4193 


.7046 


49 00 


.7547 


.6561 


1.1504 


.8693 


55 00 


.8192 


.5736 


1.4281 


.7002 


10 


.7566 


.6539 


1.1571 


.8642 


10 


.8208 


.5712 


1.4370 


.6959 


20 


.7585 


.6517 


1.1640 


.8591 


20 


.8225 


.5688 


1.4460 


.6916 


30 


.7604 


.6494 


1.1708 


.8541 


30 


.8241 


.5664 


1.4550 


.6873 


40 


.7623 


.6472 


1.1778 


.8491 


40 


.8258 


.5640 


1.4641 


.6830 


50 


.7642 


.6450 


1.1847 


.8441 


50 


.8274 


.5616 


1.4733 


.6787 


50 00 


.7660 


.6428 


1.1918 


.8391 


56 00 


.8290 


.5592 


1.4826 


.6745 


10 


.7679 


.6406 


1.1988 


.8342 


10 


.8307 


.5568 


1.4919 


.6703 


20 


.7698 


.6383 


1.2059 


.8292 


20 


.8323 


.5544 


1.5013 


.6661 


30 


.7716 


.6361 


1.2131 


.8243 


30 


.8339 


.5519 


1.5108 


.6619 


40 


.7735 


.6338 


1.2203 


.8195 


40 


.8355 


.5495 


1.5204 


.6577 


50 


.7753 


.6316 


1.2276 


.8146 


50 


.8371 


.5471 


1.5301 


.6536 


51 00 


.7771 


.6293 


1.2349 


.8098 


57 00 


.8387 


.5446 


1.5399 


.6494 


10 


.7790 


.6271 


1.2423 


.8050 


10 


.8403 


.5422 


1.5497 


.6453 


20 


.7808 


.6248 


1.2497 


.8002 


20 


.8418 


.5398 


1.5597 


.6412 


30 


.7826 


.6225 


1.2572 


.7954 


30 


.8434 


.5373 


1.5697 


.6371 


40 


.7844 


.6202 


1.2647 


.7907 


40 


.8450 


.5348 


1.5798 


.6330 


50 


.7862 


.6180 


1.2723 


.7860 


50 


.8465 


.5324 


1.5900 


.6289 


52 00 


.7880 


.6157 


1.2799 


.7813 


58 00 


.8480 


.5299 


1.6003 


.6249 


10 


.7898 


.6134 


1.2876 


.7766 


10 


.8496 


.5275 


1.6107 


.6208 


20 


.7916 


.6111 


1.2954 


.7720 


20 


.8511 


.5250 


1.6212 


.6168 


30 


.7934 


.6088 


1.3032 


.7673 


30 


.8526 


.5225 


1.6319 


.6128 


40 


.7951 


.6065 


1.3111 


.7627 


40 


.8542 


.5200 


1.6426 


.6088 


50 

N 

53 00 


.7969 


.6041 


1.3190 


.7581 


50 


.8557 


.5175 


1.6534 


.6048 


.7986 


.6018 


1.3270 


.7536 


59 00 


.8572 


.5150 


1.6643 


.6009 


10 


.8004 


.5995 


1.3351 


.7490 


10 


.8587 


.5125 


1.6753 


.5969 


20 


.8021 


.5972 


1.3432 


.7445 


20 


.8601 


.5100 


1.6864 


.5930 


30 


.8039 


.5948 


1.3514 


.7400 


30 


.8616 


.5075 


1.6977 


.5890 


40 


.8056 


.5925 


1.3597 


.7355 


40 


.8631 


.5050 


1.7090 


.5851 


50 


.8073 


.5901 


1.3680 


.7310 


50 


.8646 


.5025 


1.7205 


.5812 



TABLES 



87 



NATURAL SINES, COSINES, TANGENTS, AND COTANGENTS 
(Advanced by 10 Minutes) 



Degrees 
Minutes 


8 
m 


■ 


a 

B 

to 

a 


a 

V 

M 

a 

03 
O 

U 


Degrees 
Minutes 


© 
a 
m 


1 


i 


d 

5 
M 

a 

a 

O 

o 


60 00 
10 
20 


.8660 
.8675 
.8689 


.5000 
.4975 
.4950 


1.7321 
1.7437 
1.7556 


.5774 
.5735 
.5696 


66 00 
10 
20 


.9135 
.9147 
.9159 


.4067 
.4041 
.4014 


2.2460 
2.2637 
2.2817 


.4452 
.4417 
.4383 


30 
40 
50 


.8704 

.8718 
.8732 


.4924 
.4899 

.4874 


1.7675 
1.7796 
1.7917 


.5658 
.5619 
.5581 


30 
40 
50 


.9171 
.9182 
.9194 


.3987 
.3961 
.3934 


2.2998 
2.3183 
2.3369 


.4348 
.4314 
.4279 


61 00 
10 
20 


.8746 
.8760 
.8774 


.4848 
.4823 
.4797 


1.8040 
1.8165 
1.8291 


.5543 
.5505 
.5467 


67 00 
10 
20 


.9205 
.9216 
.9228 


.3907 
.3881 
.3854 


2.3559 
2.3750 
2.3945 


.4245 
.4210 
.4176 


30 
40 
50 


.8788 
.8802 
.8816 


.4772 
.4746 
.4720 


1.8418 
1.8546 
1.8676 


.5430 
.5392 
.5354 


30 
40 
50 


.9239 
.9250 
.9261 


.3827 
.3800 
.3773 


2.4142 
2.4342 
2.4545 


.4142 
.4108 
.4074 


62 00 
10 
20 


.8829 
.8843 
.8857 


.4695 
.4669 
.4643 


1.8807 
1.8940 
1.9074 


.5317 
.5280 
.5243 


68 00 
10 
20 


.9272 
.9283 
.9293 


.3746 
.3719 
.3692 


2.4751 

2.4960 
2.5172 


.4040 
.4006 
.3973 


30 
40 
50 


.8870 
.8884 
.8897 


.4617 
.4592 
.4566 


1.9210 
1.9347 
1.9486 


.5206 
.5169 
.5132 


30 
40 
50 


.9304 
.9315 
.9325 


.3665 
.3638 
.3611 


2.5386 
2.5605 
2.5826 


.3939 
.3906 
.3872 


63 00 
10 
20 


.8910 
.8923 
.8936 


.4540 
.4514 

.4488 


1.9626 
1.9768 
1.9912 


.5095 
.5059 
.5022 


69 00 
10 

20 


.9336 
.9346 
.9356 


.3584 
.3557 
.3529 


2.6051 
2.6279 
2.6511 


.3839 
.3805 
.3772 


30 
40 
50 


.8949 
.8962 
.8975 


.4462 
.4436 
.4410 


2.0057 
2.0204 
2.0353 


.4986 
.4950 
.4913 


30 
40 
50 


.9367 
.9377 
.9387 


.3502 
.3475 
.3448 


2.6746 
2.6985 
2.7228 


.3739 
.3706 
.3673 


64 00 

10 
20 


.8988 
.9001 
.9013 


.4384 
.4358 
.4331 


2.0503 
2.0655 
2.0809 


.4877 
.4841 
.4806 


70 00 
10 
20 


.9397 
.9407 
.9417 


.3420 
.3393 
.3365 


2.7475 
2.7725 
2.7980 


.3640 
.3607 
.3574 


30 
40 
50 


.9026 

.9038 
.9051 


.4305 
.4279 
.4253 


2.0965 
2.1123 
2.1283 


.4770 
.4734 

.4699 


30 
40 
50 


.9426 
.9436 
.9446 


.3338 
.3311 
.3283 


2.8239 
2.8502 
2.8770 


.3541 
.3508 
.3476 


65 00 
10 
20 


.9063 
.9075 
.9088 


.4226 
.4200 
.4173 


2.1445 
2.1609 
2.1775 


.4663 
.4628 
.4592 


71 00 
10 
20 


.9455 
.9465 
.9474 


.3256 
.3228 
.3201 


2.9042 
2.9319 
2.9600 


.3443 
.3411 
.3378 


30 
40 
50 


.9100 
.9112 
.9124 


.4147 
.4120 
.4094 


2.1943 
2.2113 
2.2286 


.4557 
.4522 

.4487 


30 
40 
50 


.9483 
.9492 
.9502 


.3173 
.3145 
.3118 


2.9887 
3.0178 
3.0475 


.3346 
.3314 
.3281 



88 



PLANE TRIGONOMETRY 



NATURAL SINES, COSINES, TANGENTS, AND COTANGENTS 

(Advanced by 10 Minutes) 



Degrees 
Minutes 




1 

o 


I 

! 


a 
a> 
bfl 

c3 

t 

o 


Degrees 
Minutes 


1 


8 
1 


1 

§ 


a 



a 

08 
| 


72 00 
10 
20 


.9511 
.9520 
.9528 


.3090 
.3062 
.3035 


3.0777 
3.1084 
3.1397 


.3249 
.3217 
.3185 


78 00 
10 
20 


.9781 
.9787 
.9793 


.2079 
.2051 
.2022 


4.7046 
4.7729 
4.8430 


.2126 
.2095 
.2065 


30 
40 
50 


.9537 
.9546 
.9555 


.3007 
.2979 
.2952 


3.1716 
3.2041 
3.2371 


.3153 
.3121 
.3089 


30 
40 
50 


.9799 
.9805 
.9811 


.1994 
.1965 
.1937 


4.9152 
4.9894 
5.t)658 


.2035 
.2004 
.1974 


73 00 

10 
20 


.9563 
.9572 
.9580 


.2924 
.2896 
.2868 


3.2709 
3.3052 
3.3402 


.3057 
.3026 
.2994 


79 00 

10 
20 


.9816 

.9822 
.9827 


.1908 
.1880 
.1851 


5.1446 
5.2257 
5.3093 


.1944 
.1914 
.1883 


30 
40 
60 


.9588 
.9596 
.9605 


.2840 
.2812 

.2784 


3.3759 
3.4124 
3.4495 


.2962 
.2931 
.2899 


30 
40 
50 


.9833 
.9838 
.9843 


.1822 
.1794 
.1765 


5.3955 
5.4845 
5.5764 


.1853 
.1823 
.1793 


74 00 

10 
20 


.9613 
.9621 
.9628 


.2756 

.2728 
.2700 


3.4874 
3.5261 
3.5656 


.2867 
.2836 
.2805 


80 00 
10 
20 


.9848 
.9853 
.9858 


.1736 
.1708 
.1679 


5.6713 
5.7694 
5.8708 


.1763 
.1733 
.1703 


30 

40 
50 


.9636 
.9644 
.9652 


.2672 
.2644 
.2616 


3.6059 
3.6470 
3.6891 


.2773 
.2742 
.2711 


30 
40 
50 


.9863 
.9868 
.9872 


.1650 
.1622 
.1593 


5.9758 
6.0844 
6.1970 


.1673 
.1644 
.1614 


75 00 
10 
20 


.9659 
.9667 
.9674 


.2588 
.2560 
.2532 


3.7321 
3.7760 
3.8208 


.2679 
.2648 
.2617 


81 00 
10 
20 


.9877 
.9881 
.9886 


.1564 
.1536 
.1507 


6.3138 
6.4348 
6.5606 


.1584 
.1554 
.1524 


30 
40 
50 


.9681 

.9689 
.9696 


.2504 
.2476 
.2447 


3.8667 
3.9136 
3.9617 


.2586 
.2555 
.2524 


30 
40 
50 


.9890 
.9894 
.9899 


.1478 
.1449 
.1421 


6.6912 
6.8269 
6.9682 


.1495 
.1465 
.1435 


76 00 
10 
20 


.9703 
.9710 
.9717 


.2419 
.2391 
.2363 


4.0108 
4.0611 
4.1126 


.2493 
.2462 
.2432 


82 00 
10 
20 


.9903 
.9907 
.9911 


.1392 
.1363 
.1334 


7.1154 

7.2687 
7.4287 


.1405 
.1376 
.1346 


30 
40 
50 


.9724 
.9730 
.9737 


.2334 
.2306 

.2278 


4.1653 
4.2193 
4.2747 


.2401 
.2370 

.2339 


30 
40 
50 


.9914 
.9918 
.9922 


.1305 
.1276 
.1248 


7.5958 
7.7704 
7.9530 


.1317 
.1287 
.1257 


77 00 
10 
20 


.9744 
.9750 
.9757 


.2250 
.2221 
.2193 


4.3315 
4.3897 
4.4494 


.2309 

.2278 
.2247 


83 00 
10 
20 


.9925 
.9929 
.9932 


.1219 
.1190 
.1161 


8.1443 
8.3450 
8.5555 


.1228 
.1198 
.1169 


30 
40 
50 


.9763 
.9769 
.9775 


.2164 
.2136 
.2108 


4.5107 
4.5736 
4.6382 


.2217 
.2186 
.2156 


30 
40 
50 


.9936 
.9939 
.9942 


.1132 
.1103 
.1074 


8.7769 
9.0098 
9.2553 


.1139 
.1110 
.1080 



TABLES 



89 



NATURAL SINES, COSINES, TANQENTS, AND COTANQENTS 
(Advanced by 10 Minutes) 



Degrees 
Minutes 


1 


m 
a 

1 

O 


1 

§ 


-*> 

a 

V 

5 

a 

"5 


Degrees 
Minutes 


a 


00 

1 


1 

B 

a 
3 


§ 

M 

a 

e3 
* 


84 00 


.9945 


.1045 


9.5144 


.1051 


87 00 


.9986 


.0523 


19.081 


.0524 


10 


.9948 


.1016 


9.7882 


.1022 


10 


.9988 


.0494 


20.206 


.0495 


20 


.9951 


.0987 


10.078 


.0992 


20 


.9989 


.0465 


21.470 


.0466 


30 


.9954 


.0958 


10.385 


.0963 


30 


.9990 


.0436 


22.904 


.0437 


40 


.9957 


.0930 


10.712 


.0934 


40 


.9992 


.0407 


24.542 


.0407 


50 


.9959 


.0901 


11.059 


.0904 


50 


.9993 


.0378 


26.432 


.0378 


85 00 


.9962 


.0872 


11.430 


.0875 


88 00 


.9994 


.0349 


28.636 


.0349 


10 


.9964 


.0843 


11.826 


.0846 


10 


.9995 


.0320 


31.242 


.0320 


20 


.9967 


.0814 


12.251 


.0816 


20 


.9996 


.0291 


34.368 


.0291 


30 


.9969 


.0785 


12.706 


.0787 


30 


.9997 


.0262 


38.188 


.0262 


40 


.9971 


.0756 


13.197 


.0758 


40 


.9997 


.0233 


42.964 


.0233 


50 


.9974 


.0727 


13.727 


.0729 


50 


.9998 


.0204 


49.104 


.0204 


86 00 


.9976 


.0698 


14.301 


.0699 


89 00 


.9999 


.0175 


57.290 


.0175 


10 


.9978 


.0669 


14.924 


.0670 


10 


.9999 


.0145 


68.750 


.0145 


20 


.9980 


.0640 


15.605 


.0641 


20 


.9999 


.0116 


85.940 


.0116 


30 


.9981 


.0610 


16.350 


.0612 


30 


one 


.0087 


114.59 


.0087 


40 


.9983 


.0581 


17.169 


.0582 


40 


one 


.0058 


171.89 


.0058 


50 


.9985 


.0552 


18.075 


.0553 


50 


one 


.0029 


343.77 


.0029 



